Wednesday

April 16, 2014

April 16, 2014

Number of results: 30,438

**Calculus Infinite Geometric Series**

You have missed some basic material. if r is between -1 and 1, the series converges, and you can get an infinite series sum If r is greater than 1, or less than -1, the series gets larger and larger, and diverges, and you cannot get a sum. Consider the series as 10+100+1000+...
*Monday, September 19, 2011 at 8:27pm by bobpursley*

**Calculus**

a) Find the Taylor series associated to f(x) = x^-2 at a = 1. Be sure to show the general term of the series. b) Find the radius of convergence of the series. c)Use Lagrange's Remainder Theorem to prove that for x in the interval of convergence with x > 1; the power series ...
*Monday, August 8, 2011 at 8:30pm by laura*

**calculus**

im confused. I was taught that if it is an alternating series then you can only use the alternating series test to see if it was diverging or converging. And i did the alternating series test and all the conditions were met for each series seperately.
*Sunday, July 31, 2011 at 8:12pm by amber*

**Calculus**

Consider the infinite series of the form: (+/-)3(+/-)1(+/-)(1/3)(+/-)(1/9)(+/-)(1/27)(+/-)...(+/-)(1/3^n)(+/-)... (A) Find x and y from: x(</=)(+/-)3(+/-)1(+/-)(1/3)(+/-)...(</=)y. (B) Can you choose the signs to make the series diverge? (C) Can you choose the signs to ...
*Wednesday, April 21, 2010 at 5:51pm by Megan*

**calculus**

Consider the infinite series of the form: (+/-)3(+/-)1(+/-)(1/3)(+/-)(1/9)(+/-)(1/27)(+/-)...(+/-)(1/3^n)(+/-)... (A) Find x and y from: x(</=)(+/-)3(+/-)1(+/-)(1/3)(+/-)...(</=)y. (B) Can you choose the signs to make the series diverge? (C) Can you choose the signs to ...
*Thursday, April 22, 2010 at 2:23am by Megan*

**calculus**

Consider the infinite series of the form: (+/-)3(+/-)1(+/-)(1/3)(+/-)(1/9)(+/-)(1/27)(+/-)...(+/-)(1/3^n)(+/-)... (A) Find x and y from: x(</=)(+/-)3(+/-)1(+/-)(1/3)(+/-)...(</=)y. (B) Can you choose the signs to make the series diverge? (C) Can you choose the signs to ...
*Thursday, April 22, 2010 at 9:54pm by Megan*

**Calculus**

Determine the following about the series. Indicate the test that was used and justify your answer. Sigma (lower index n = 1; upper index infinity) [sin((2n-1)pi/2)]/n A. The series diverges B. The series converges conditionally. C. The series converges absolutely. D. It cannot...
*Monday, February 11, 2013 at 6:06pm by Jenny*

**College Calculus**

Suppose the series An (from n=1 to INF) is known to be convergent. Prove that series 1/(An) (from n=1 to INF) is a divergent series. I have no idea what to do... please help!
*Monday, March 24, 2008 at 8:05pm by Mandy*

**College Calculus (Binomial Series)**

Expand f(x) = (x+x^2)/((1-x)^3) as a power series and use it to find the sum of series (SUM from n=1 to infinity) (n^2)/(2^n) PLEASE HELP.
*Tuesday, April 8, 2008 at 9:13pm by Andrew*

**calculus**

With power series, is an endpoint convergent if you plug it back into the original series, and get an alternating series that is conditionally convergent?
*Wednesday, September 22, 2010 at 6:12pm by Bob*

**Calculus**

By recognizing each series below as a Taylor series evaluated at a particular value of x, find the sum of each convergent series. A) 1+5 + (5^2)/(2!)+(5^3)/(3!)+(5^4)/(4!)+...+ (5^k)/(k!)+...= B) 1-(2^2)/(2!)+(2^4)/(4!)-(2^6)/(6!)+...+((-1)^(k)2^(2k))/((2k)!) +...=
*Tuesday, June 18, 2013 at 10:24pm by Cole*

**Calculus**

How can I prove this series alternating series converges(this is the answer)? (-1)^2*(2/3)^n I tried it this way: an = (2/3)^n, then i just broke it down. 2^n/(3^n) and i took the ratio of it and got 2/3 which does not equal to one which would mean the series diverges.. but ...
*Friday, July 26, 2013 at 11:23pm by Isaac*

**Calculus**

Obtain the MacLaurin series for 1/(2-x) by making an appropriate substitution into the MacLaurin series for 1/(1-x). ------------ The MacLaurin series for 1/(1-x) = Σ x^k I substitue (x-1) in for x, because 1/(2-x) = 1/(1-(x-1)) Making the same substitution in the ...
*Thursday, June 25, 2009 at 7:55pm by Sean*

**Calculus**

Find a power series, centered @ x=0, for function f(x)=x/(1+2x). I know this is a maclaurin series, but my work doesn't get the right answer. Can you please show steps? Also,do all power series start with a 1, as in (1+2x+4x^2+...)? Thanks in advance!
*Thursday, March 8, 2012 at 8:21pm by W*

**Calculus**

Find a power series, centered @ x=0, for function f(x)=x/(1+2x). I know this is a maclaurin series, but my work doesn't get the right answer. Can you please show steps? Also,do all power series start with a 1, as in (1+2x+4x^2+...)? Thanks in advance!
*Thursday, March 8, 2012 at 9:39pm by W*

**Calculus, series**

That is a sequence, not a series. A sequence is an ordered list of terms. A series is a sum of terms, which may be terms in a sequence. so 1,5,1,5,1,5..... is a sequence but 1+5+1+5+1+5..... would be a series the pattern of your sequence could be said to be alternating 1 and 5
*Wednesday, January 9, 2008 at 6:46pm by Damon*

**calculus**

the question asks to use power series though. So i found the power series of ln(1-x) = Series of nx^n-1 and for e^x is the series of x^n/n! i then found the first three terms of each ln(1-x) = 1 + 2x +3x^2 e^x = 1 + x^2/2! + x^3?3! after multiply them i got 1 + 2x + (X^2/2!+3x...
*Monday, August 1, 2011 at 9:15am by amber*

**CALCULUS**

find the radius and interval of convergence for the series the series from n=0 to infinity of ((-1)^n*x^n)/(n+1)
*Tuesday, March 4, 2008 at 9:13pm by jane*

**Calculus, series**

I cannot figure this out! What is the series (the Pattern) of this sequence? {1,5,1,5,1,5,1,5.......}
*Wednesday, January 9, 2008 at 6:46pm by Candace*

**calculus**

how do you find the sum of a series? for example: the series from n=0 to infinity of (-1)^n/n!??? thanks
*Thursday, March 20, 2008 at 8:54pm by sarah*

**College Calculus (Binomial Series)**

Taylor series is as follows: f(x)=f(a)+f'(a)(x-a)/1!+f''(a)(x-a)^2/2!+f'''(a)(x-a)^3/3!+...
*Tuesday, April 8, 2008 at 9:13pm by FredR*

**CALCULUS-URGENT**

find the radius and interval of convergence for the series the series from n=0 to infinity of ((-1)^n*x^n)/(n+1)
*Tuesday, March 4, 2008 at 10:05pm by jane*

**calculus**

From Taylor (Maclaurin series if starting at x =0) series f(x)=f(0) + f'(0)(x/1!) + f""(0)x^2/2!).... here f(x) = e^x f(0) = 1 the derivatives of e^x are all e^x which is 1 at x = 0 so e^x = 1 + 1(x/1!) +1(x^2/2! etc
*Monday, May 31, 2010 at 8:19pm by Damon*

**Calculus**

does infinite power series -1/n diverge? (i think so, because it is just the negative of the harmonic series)...? thank you!
*Saturday, October 20, 2012 at 10:13am by Nikky*

**calculus**

what does this series, if it converges, converge to? the series from n=1 to infinity of (2^n + (-1)^n)/3^n
*Monday, March 17, 2008 at 12:30am by sarah*

**calculus**

"If the sequence An converges to 0, and each An is smaller than An-1 (i.e. the sequence An is monotone decreasing), then the series converges." In this case, An = 1 / (3n + 1) The limit as n->infinity (An) = 0 A1 = 1 / 4 A2 = 1 / 7 Therefore, by the Alternating Series Test ...
*Thursday, February 28, 2008 at 10:55pm by Dan*

**calculus III**

All of your series A, B and C are the MacLaurin series expansion for ln(1 +x), where x = 1. You are just doing the summation in a different order. The exact sum for the infinite series is ln(2) = 0.69315... I don't see the point of adding up 30 terms in different orders, but ...
*Wednesday, October 13, 2010 at 12:01am by drwls*

**CALCULUS-URGENT- no one will respond!!!**

find the radius and interval of convergence for the series the series from n=0 to infinity of ((-1)^n*x^n)/(n+1)
*Wednesday, March 5, 2008 at 12:08am by jane*

**Calculus 2**

Hello, I don't know what test to use for this series: Determine the sum of the following series: inf E n=1 (2^n + 9^n) / 12^n thank you!
*Friday, March 22, 2013 at 6:20pm by Tom*

**calculus**

state the power series of an appropriate familiar function and use it to calculate the power series of the given function. give answer in sigma notation. All power series have center at 0. f(x)=(e^(-x)^2) -1+(x^2) and g(x)=x cos (x/square root 3)
*Monday, April 20, 2009 at 5:05pm by pervaiz*

**calculus**

There is no single method to do them all. In the case of that series, it looks like the Taylor series for e^x when x = -1, so the answer is 1/e = 0.35935....
*Thursday, March 20, 2008 at 8:54pm by drwls*

**calculus**

Use division of power series to find the first three terms of the Maclaurin series for y = sec x.
*Monday, August 1, 2011 at 7:17pm by ami*

**calculus**

i have to determine whether the series is convergent, and if, find the sum the series is from k=1 to infinity of 2/((k+1)(k+3)) I got 5/6 as my answer and didn't know if it was right...
*Thursday, February 21, 2008 at 6:15pm by sarah*

**Calculus**

Show that the following series is absolutely convergent: Summation from 1 to infinity: [(-1)^n * (n+1) * 3^n]/ [2^(2n+1)] I've done the ratio test and replaced n in this series with n+1. I get 3/4 in the end, which is less than 1, which confirms that the series is abs. ...
*Saturday, March 24, 2012 at 6:17pm by Anonymous*

**calculus**

Use multiplication of power series to find the first three non-zero terms of the Maclaurin series of e^x ln(1 − x).
*Monday, August 1, 2011 at 9:15am by amber*

**Calculus**

I assume you mean (-1)^n * (2/3)^n This is just a geometric series with r = -2/3 So, if you start with n=0, the sequence starts with 1, and Sum = 1/(1-r) = 1/(1+2/3) = 3/5 In an alternating series, if the ratio |r| < 1 it converges.
*Friday, July 26, 2013 at 11:23pm by Steve*

**calculus**

You need to insert te upper limit in the series, the lower limit and then subtract. However, in this that doesn't work as the series doesn't converge at the upper limit. What you need to do is to write down the series that converges for x larger than one and the series that ...
*Wednesday, March 19, 2008 at 6:52am by Count Iblis*

**Calculus - Taylor #2**

Find the Taylor series for f(x) centered at the given value of 'a'. (Assume that 'f' has a power series expansion. Do not show that Rn(x)-->0.) f(x) = x3, a = -1 and what i've done so far: f (x) = x^3 f ' (x) = 3x^2 f '' (x) = 6x^1 f ''' (x) = 6x f (-1) = -1 f ' (-1) = 3 f...
*Saturday, July 28, 2007 at 1:02pm by COFFEE*

**calculus**

Consider ∞ ∑ [(3k+5)/(kČ-2k)]ᵖ, for each p ∈ ℝ. k=3 Show this series { converges if p > 1 { diverges if p ≤ 1 Hint: Determine the known series whose terms past the second give an approximate match for the terms of this series. This series...
*Saturday, August 6, 2011 at 10:58am by Bernie*

**calculus**

I need help understanding how the series of e derives into the exponential series using the binomial theorem. Here is a link to a pic of a page in my book, regarding the exponential series: ht tp://i46.tiny pic(.)(com)/qz0oat . jpg (remove parentheses and spaces) A couple of ...
*Saturday, June 5, 2010 at 8:56pm by Jimmy*

**calculus**

Log(e^2 + 1/10) = Log(e^2) + Log[1+e^(-2)/10] We can compute this using the series expansion: Log(1+x) = x - x^2/2 + x^3/3 - x^4/4 + ... But we can speed up the convergence of the series as follows. Replacing x by -x in the series gives: Log(1-x) = -x - x^2/2 - x^3/3 - x^4/4...
*Sunday, June 16, 2013 at 6:51pm by Count Iblis*

**calculus**

It is not clear without sufficient parentheses what the expression really is. I assume it to be: Sum((-1)^n * (3/3^n)) for n=0 -> ∞ This is an alternating geometric series. (9/2) is the correct sum for the geometric series (non-alternating). Write out the first few ...
*Tuesday, September 7, 2010 at 8:10pm by MathMate*

**Calculus**

If you have a geometric alternating series, and you prove that the series is converging by doing geometric series test, and NOT alternating series test, then does that allow you to say that the series converges ABSOLUTELY? Or should you do alternate series test also to say ...
*Sunday, March 4, 2007 at 7:29pm by Jin*

**calculus**

test the series for convergence or divergence the series from n=1 to infinity of 1/(arctan(2n)) I again didn't know what test to use
*Sunday, March 16, 2008 at 2:59pm by lila*

**Calculus**

Consider an infinite series of the form (+-)3(+-)1(+-)1/3(+-)1/9(+-)1/27(+-)....(+-)1/3^n(+-)... The number 3,1, etc. are given but you will decide what the signs should be. a)Can you choose the signs to make the series diverge? B)Can you choose the signs to make the series ...
*Monday, November 12, 2012 at 4:32pm by oci*

**calculus**

test the series for convergence or divergence the series from n=0 to infinity of (x^2+1)/(x^3+1) I said that due to the limit comparison test this converges at 1
*Sunday, March 16, 2008 at 2:57pm by lila*

**calculus**

test the series for convergence or divergence the series from n=0 to infinity of sin (2^-x) I wasn't sure what test to use to see if this was or wasn't convergent
*Sunday, March 16, 2008 at 2:58pm by lila*

**calculus**

for each series determine if the series is absolutely convergent and convergent the sum from 0 to infinity of (-1)^n/(the square root of (n+1))
*Sunday, March 2, 2008 at 6:01pm by sarah*

**calculus**

determine whether the series is convergent if so find sum it is the sum from k=1 to infinity of ((-1)^k)/(3^(k+1)) i found this series to be geometric where a=-1/9 and r=1/3 my answer was converges to 1/6
*Thursday, February 21, 2008 at 7:56pm by sarah*

**College Calculus**

If the series is convergent then you know that An tends to zero, therefore 1/An tends to infinity and thus the second series cannot converge
*Monday, March 24, 2008 at 8:05pm by Count Iblis*

**Calculus**

1 + .865 + .865^2 +.865^3 ...... this is a geometric series with the first term one and every successive term multiplied by r =.865 (approx) If r <1 then the geometric series converges. for the series a + a r + a r^2 + a r^3 ----- a r^oo the sum is S = a/(1-r) here a = 1 ...
*Monday, March 4, 2013 at 3:47pm by Damon*

**calculus**

Another problem: determine whether the series is convergent if so find sum it is the sum from k=1 to infinity of ((-1)^k)/(3^(k+1)) i found this series to be geometric where a=-1/9 and r=1/3 my answer was converges to 1/6
*Thursday, February 21, 2008 at 6:33pm by sarah*

**calculus**

test the series for convergence or divergence using the alternating series test the sum from n=1 to infinity of (-1)^n/(3n+1) I said it converges, is this true?
*Wednesday, February 27, 2008 at 3:22pm by sarah*

**calculus**

test the series for convergence or divergence using the alternating series test the sum from n=1 to infinity of (-1)^n/(3n+1) I said it converges, is this true?
*Thursday, February 28, 2008 at 10:55pm by laura*

**calculus**

It looks like that can be written as the sum of two series: (2/3)^n and (-1/3)^n Use the fact that the sum of the series from 1 to infinity r^n = 1/(1-r) , for |r|<1 That makes the limit 1/(1/3) + 1/(4/3) = 3 3/4
*Monday, March 17, 2008 at 12:30am by drwls*

**calculus**

It depends on the radius of convergence. If the series has a radius of convergence of ∞ such as the series for sin(x), cos(x), and exp(x), you can evaluate f(x) where x equals any defined value, even though the series are expanded about an arbitrary value x0. In the ...
*Tuesday, March 15, 2011 at 8:16pm by MathMate*

**calculus**

So, you're saying that when you separate an alternating series into two (odd and even) then the (-1)^n-1 goes away? And why did you make S1 negative, but not S2. And do i need to find the sum for S2 to answer the question? Because i only know how to find the sum of geometric ...
*Sunday, July 31, 2011 at 8:12pm by amber*

**Calculus III**

1. Apparently you mean the sum from n=1 to n=infinity of 1/(6n^2) = 1/(6*1) + 1/(6*4) + 1/6*9) = (1/6)(1 + 1/4 + 1/9 + 1/16 + ...)] This is not a geometric series. The ratio of successive terms is not a constant. 2. 0.6 + 0.6^2 + 0.6^3 + ... = 0.6(1 + 0.6 + 0.6^2 + ...] = 0.6...
*Monday, December 10, 2007 at 11:06pm by drwls*

**calculus**

does this series converge, and if so is it absolutely convergent? the series from n=1 to infinity of ((-1)^*n+1))/n^4 I found that by the ratio test it was inconclusive, so no abs. conv is this right? and how do i know if it is simply convergent?
*Monday, March 17, 2008 at 7:00am by matt*

**calculus**

does this series converge, and if so is it absolutely convergent? the series from n=1 to infinity of ((-1)^*n+1))/n^4 I found that by the ratio test it was inconclusive, so no abs. conv is this right? and how do i know if it is simply convergent?
*Monday, March 17, 2008 at 7:14am by matt*

**Calculus - Series**

so I realized it's not using the derivative that we find if a sequenec is ascending/descending, but rather comparing the series. Nonetheless, the sequence is still descending.
*Wednesday, September 19, 2012 at 6:02pm by Paul*

**calculus**

Where did the exponential series come from? 1 + x + x^2/2! + x^3/3!... Where did that number come from? and how is it used to get the trigonometric series?
*Monday, May 31, 2010 at 8:19pm by Miranda*

**Calculus**

If you notice that the Taylor's series expansion of e^(-x/4) = 1-x/4+x^2/32-x^3/384+x^4/6144-... is exactly the given series less the first term with x=1, so the given series is e^(-1/4)-1 = -0.2212 to four places. If you sum term by term, you just need to sum until the next ...
*Tuesday, May 8, 2012 at 7:07pm by MathMate*

**Excel Help**

3) __________ are used to compare sets of data in one chart. Time series Multiple data series Relative series Comparison series Is multiple data series
*Sunday, July 12, 2009 at 10:29pm by Bryan*

**calculus**

As k becomes large, e^-k becomes much less than 1, and sin(e^-k) approaches e^-k The sum of the series 1 + 1/e + 1/e^2 converges to 1 /(1 - 1/e)= 1.582 High-order terms of the series sin(e^-k) will behave similarly, but the sum of the entire series will be somethat less than 1...
*Tuesday, February 26, 2008 at 11:08pm by drwls*

**calculus**

How do you differentiate the maclaurin series for 1/(1-x) twice to find the maclaurin series of 1/(1-x)^3.
*Monday, April 26, 2010 at 5:16pm by Paul*

**Calculus Derivative- Taylor Series?**

let f(x)= x/x-1 find f'(x) f ''(x) and a formula for f ^ (n) * x. I found the first and second derivatives but do not know how to make a general equation for this. I have not learnt the Taylor or Maclaurin Series either. Thank you.
*Sunday, March 10, 2013 at 5:37pm by Lisa*

**Calculus**

If S1 = 0.7 and S2 = 2.1 in geometric series, what would the sum of the first 12 terms in the series be? I tried doing this, and I got 1.16 or something :S How exactly do I do this? Please tell me what formula to use. I was using the Sn = a(rn-1)/r-1
*Tuesday, September 20, 2011 at 8:16pm by Anonymous*

**Calculus**

that doesn't mean it's convergent, necessarily. but i found it is actually convergent because if you divide by 1/2^n, which is also a positive term series, you get 9, which means the series is convergent.
*Thursday, December 4, 2008 at 3:01pm by Shaniquaa*

**calculus**

use the power series to estimate the series: from 0 to 4 of ln(1+x)dx with absolute value of the error less than .0001/ Give your estimate of the integral as well as a bound on the error. I found the 'terms' in the series to be: x-(1/2)x^2+(1/3)x^3-(1/4)x^4...... with a radius...
*Tuesday, March 18, 2008 at 10:42pm by meg*

**calculus**

use the power series to estimate the series: from 0 to 4 of ln(1+x)dx with absolute value of the error less than .0001/ Give your estimate of the integral as well as a bound on the error. I found the 'terms' in the series to be: x-(1/2)x^2+(1/3)x^3-(1/4)x^4...... with a radius...
*Wednesday, March 19, 2008 at 6:52am by meg*

**Calculus - - Reiny**

If S1 = 0.7 and S2 = 2.1 in geometric series, what would the sum of the first 12 terms in the series be? I tried doing this, and I got 1.16 or something :S How exactly do I do this? Please tell me what formula to use. I was using the Sn = a(rn-1)/r-1
*Tuesday, September 20, 2011 at 8:47pm by Anonymous*

**calculus**

Is ((-1)^*n+1)) supposed to mean [(-1)^(n+1)] ? If so, an infinite series of positive 1/n^4 terms is convergent, based on the integral test. A similar series with alternating + or - terms of the same magnitude must therefore also converge.
*Monday, March 17, 2008 at 7:00am by drwls*

**calculus**

determine if the series is absolutely convergent and convergent the sum from n=1 to infinity of sin(n^2)/n^2 what series test should I use and how? the ratio test?
*Sunday, March 2, 2008 at 6:08pm by sarah*

**calculus**

determine whether the series converges of diverges the sum from k=2 to infinity of (the square root of (ln(k)))/k I said that because you can't integrate the series (goes to infinity) it diverges
*Tuesday, February 26, 2008 at 1:44am by sarah*

**calculus**

determine whether the series converges of diverges the sum from k=2 to infinity of (the square root of (ln(k)))/k I said that because you can't integrate the series (goes to infinity) it diverges is this true?
*Tuesday, February 26, 2008 at 8:37pm by sarah*

**Calculus 2**

Find the radius of convergence and interval of convergence of the series. n=2 series to infinity (-1)^n * x^n+6/n+7 R= ? I= (or[ , ]or) How do i do this?? ( means convergent and [ means divergent.
*Monday, November 5, 2012 at 12:50am by James*

**CALCULUS**

find the radius and interval of convergence for the series the series from n=1 to infinity of ((-1)^(n+1)*x^n)/n! I did the ratio test so I had the Lim as n approaches infinity of -x/(n+1), but this is 0, giving no radius, so I think I did something wrong...
*Tuesday, March 4, 2008 at 9:12pm by jane*

**Calculus 2**

The following function has a series of the form the sum from n=0 to infinity of c(subn)x^n. Calculate the coefficients c(subn) and express the power series in summation notation. f(x)=(pi*x)/(pi*x+1) Thank you so much for your help!!!!
*Wednesday, November 28, 2012 at 11:48pm by Heather*

**Good Reads? 6th and 7th grade**

They Cay Walk to Moons The Hunger Games Series Travels with Charley A Year Down Yonder Charlotte's Web Harry Potter Series Percy Jackson Series Diary of a Wimpy Kids Series Holes
*Tuesday, May 24, 2011 at 7:28pm by Elina*

**CALCULUS-URGENT**

find the radius and interval of convergence for the series the series from n=1 to infinity of ((-1)^(n+1)*x^n)/n! I did the ratio test so I had the Lim as n approaches infinity of -x/(n+1), but this is 0, giving no radius, so I think I did something wrong...
*Tuesday, March 4, 2008 at 10:05pm by jane*

**Calculus**

Consider 1 - 1/2 + 1/3 - 1/4 + ... (alternating harmonic series) 1 + 1/4 + 1/9 + 1/16 + ... = π2/6 1 + 1/2 + 1/3 + 1/4 + ... diverges (harmonic series)
*Thursday, October 20, 2011 at 10:29am by Steve*

**CALCULUS-URGENT- no one will respond!!!**

find the radius and interval of convergence for the series the series from n=1 to infinity of ((-1)^(n+1)*x^n)/n! I did the ratio test so I had the Lim as n approaches infinity of -x/(n+1), but this is 0, giving no radius, so I think I did something wrong...
*Wednesday, March 5, 2008 at 12:07am by laura*

**calculus - power series ASAP please :)**

using power series, integrate & evaluate to 4 dec. places integral from 0 to 1: sin x^2 dx i'm REALLY stuck on this. and i need help asap.. what is the inverse of "sin x^2" so that i could have it in a fraction that will fit the power series equation? and that is: (A)/(1-R) ...
*Tuesday, July 31, 2007 at 1:48pm by COFFEE*

**calculus**

The series can be considered as the sum of two series S1 & S2: for n odd: S1 =-∑ 1/sqrt(n) for n even: S2 = ∑ 1/2^n Assuming the summation is for n=1 to ∞, S2 sums to 1 and S1 diverges (by comparison with ∑1/n). The alternating series test requires two ...
*Sunday, July 31, 2011 at 8:12pm by MathMate*

**calculus**

Show that the series(−1)^n−1(bn) where bn = 1/(n^1/2) if n is odd and bn = 1/2^n if n is even, diverges. Why does the alternating series test fail?
*Sunday, July 31, 2011 at 8:12pm by amber*

**math**

hi if i wanna find the Maclaurin series from another existing series by integrating that series , do i need to include the constant of integration ?? Thanks
*Sunday, May 16, 2010 at 4:08am by William*

**math**

for this infinite series (-1)^n/n^2 if i use alternating series test to show that sequence does of a^n does not go to 0, then does this mean that this series is diverging
*Tuesday, July 27, 2010 at 7:49pm by eng*

**Arithmetic Series - - Reiny**

Consider the series defined by Sn = 3n-1 Find the first four terms of the series. How exactly do I do this?
*Monday, September 12, 2011 at 11:35pm by Anonymous*

**math**

the last two tems in a geometric series are 1080 and 6480 and the sum ofthe series is 7775 what is the first term in the series
*Sunday, December 18, 2011 at 2:29pm by Heather*

**chem**

Your question isn't clear at all as to what you are asking; however, the activity series (EMF series) arrangement is this. An element in the series will displace the ion of another element below it in the series. Therefore, Al will displace Pb^2+, Zn will displace H^+ etc.
*Tuesday, January 24, 2012 at 12:49am by DrBob222*

**calculus**

determine whether the series converges of diverges the sum from n=1 to infinity of 1/(the square root of (n^3+1)) I said that through the comparision test (comparing to 1/the square root of (n^3) the series converges
*Tuesday, February 26, 2008 at 1:43am by sarah*

**calculus**

is this correct? use the integral test to determine if this series is convergent or divergent: the series from n=2 to infinity of 1/(n*square root of (ln(n))) I said it was divergent because the integral went to infinity
*Sunday, March 16, 2008 at 6:40pm by john*

**Pre-calculus**

Which of the following series is divergent? a) 1+3(1/4)+9(1/4)^2+27(1/4)^3... b) 1+3(1/5)+9(1/5)^2+27(1/5)^3... c) 1+3(1/7)+9(1/7)^2+27(1/7)^3... d) 1+3(1/2)+9(1/2)^2+27(1/2)^3... How do you determine if a series in convergent or divergent??? The book that I have is about as ...
*Monday, October 13, 2008 at 11:55am by Lucy*

**calculus**

determine whether the series converges of diverges the sum from n=1 to infinity of 1/(the square root of (n^3+1)) I said that through the comparision test (comparing to 1/the square root of (n^3) the series converges is this true?
*Tuesday, February 26, 2008 at 8:36pm by sarah*

**Calculus**

Can someone prove (informally) the following theory: If there is a differentiable function, f, that is represented by a Taylor series, T, then the convergence interval for series T is identical to the convergence interval for the term-by-term derivative T'.
*Monday, June 22, 2009 at 1:13pm by Sean*

**Calculus III**

Which of the following series are geometric series? Find the sum if they are 1. Infinity (Summation sign) n = 1 1/6n^2 2. Infinity (Summation sign) n = 1 (0.6)^n-1
*Monday, December 10, 2007 at 11:06pm by kile*

**C++**

I suggest you talk to the teacher or the author of the website and explain that the series diverges. If it is a public site, I'd like to know the link. On the other hand, I don't know if you got the right web site, I believe your question probably started with: "Write a MAIN ...
*Friday, March 9, 2012 at 2:40pm by MathMate*

**calculus**

The first one is an arithmetic series where a=10 and d=4 Use 138 as the last term to find the number of terms in the series, then use the sum of terms formula. The second is a geometric series with a=6 and r=-2 Use 1536 to find the number of terms you have, then the sum of ...
*Monday, January 21, 2008 at 9:42pm by Reiny*

**Calculus**

This is going to be pretty hard to show as text since it would be easier for me to post a picture of the question. The question has f(x) = x/4^2 - (2x^3)/4^4 + (3x^5)/4^6 + ... . I am trying to find out the value of f(2). There is a hint to differentiate the power series ...
*Monday, October 4, 2010 at 9:22pm by Douglas*

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