The problem with these two questions is that I cannot determine the a and r. The 3rd question-I don't know what I did wrong. Thanks for the help!
Tell whether the series converges or diverges. If it converges, give its sum.
infinity
(sigma) sin^n (pi/4 + n pi)
n=0
answer: converges, 2-sqrt(2)
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1/2 + 2/3 + 3/4 + 4/5 + ... + n/(n+1)+...
even number question. don't know the answer
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infinity
(sigma) (e/pi)^n
n=1
answer: e/(e-pi)
work:
-1<e/pi<1 Therefore, converges
a=1, r=e/pi
S= 1/(1-e/pi)
=1/(pi/pi-e/pi)
=1/[(pi-e)/pi]
=pi/(pi-e) (not the same as the answer)
The second series sum cannot converge because it becomes an infinite sum of 1's, as n goes to infinity.
The third series has terms
e/pi + (e/pi)^2 + (e/pi)^3 + ...
= (e/pi)[1 + (e/pi) + (e/pi)^2+...]
= (e/pi)/(1 - e/pi)]
= e/(pi - e)
That also does not agree with the answer in your book, but I believe it is correct. My answer is positive while e/(e-pi) would be a negative sum, with terms that are all positive. That is not possible.
I answered #3 previously. (See #2)
For #1, the individual terms are
1 - (sqrt2)/2 + [(sqrt2)/2]2 - ...
= 1 /[1 + sqrt2/2]
= [1 - sqrt2/2]/(1/2)
= 2 - sqrt 2
To determine whether a series converges or diverges and find its sum, you typically need to identify the pattern of the terms in the series and apply a relevant convergence test. In the case of the first question:
1. The series is expressed as the sum of sin^n(pi/4 + n*pi) from n=0 to infinity.
2. First, try to identify the pattern of the terms in the series. Recall that sin(pi/4 + n*pi) can be simplified using the periodicity of the sine function: sin(pi/4 + n*pi) = sin(pi/4). Therefore, the terms in the series become sin^n(pi/4).
3. Notice that when n is even, sin^n(pi/4) = 1, and when n is odd, sin^n(pi/4) = sin(pi/4). Rewriting the series:
(sin^0(pi/4) + sin(pi/4) + sin^2(pi/4) + sin^3(pi/4) + ...)
4. Simplify the terms further: sin^0(pi/4) = 1 and sin(pi/4) = sqrt(2)/2.
(1 + sqrt(2)/2 + sqrt(2)/2 + sqrt(2)/2 + ...)
5. Notice that the terms alternate between 1 and sqrt(2)/2, which form a geometric sequence with a common ratio of sqrt(2)/2.
6. The series will converge if the common ratio is between -1 and 1 (exclusive). In this case, sqrt(2)/2 is indeed between -1 and 1.
7. Use the sum formula for a convergent geometric series: S = a / (1 - r), where a is the first term and r is the common ratio.
S = 1 / (1 - sqrt(2)/2)
S = 1 / (2 - sqrt(2))
S = (2 + sqrt(2)) / (2 - sqrt(2))
S = (2 + sqrt(2)) * (2 + sqrt(2)) / ((2 - sqrt(2)) * (2 + sqrt(2)))
S = (4 + 4sqrt(2) + 2) / (4 - 2)
S = (6 + 4sqrt(2)) / 2
S = 3 + 2sqrt(2) = 2 - sqrt(2)
Therefore, the series converges and its sum is 2 - sqrt(2).
For the second question regarding the series 1/2 + 2/3 + 3/4 + 4/5 + ..., it seems you haven't specified whether it is a finite series or an infinite series. If you know it's an infinite series, then you can proceed as follows:
1. Observe the pattern of the terms: each term is given by n / (n+1), where n takes on positive integer values.
2. Rewrite the series: (1/2) + (2/3) + (3/4) + (4/5) + ...
3. This series does not have a clear pattern that fits a well-known convergence test, such as a geometric or telescoping series. It may require a different approach or use of an integral test, limit comparison test, or other methods to determine convergence or divergence.
4. Without further information, it is not possible to determine whether this series converges or diverges.
For the third question:
1. The series is expressed as the sum of (e/pi)^n from n=1 to infinity.
2. Notice that (e/pi)^n can also be written as e^n / pi^n.
3. Rewrite the series in a more recognizable form: (e/pi)^1 + (e/pi)^2 + (e/pi)^3 + ...
4. This series appears to be a geometric series with a common ratio of e/pi.
5. To determine if the series converges, we need to check if the common ratio is between -1 and 1 (exclusive).
6. In this case, the value of e/pi is approximately 1.1557, which is between -1 and 1. Therefore, the series converges.
7. Use the formula for the sum of a convergent geometric series: S = a / (1 - r), where a is the first term and r is the common ratio.
S = (e/pi)^1 / (1 - e/pi)
S = e/pi / (1 - e/pi)
S = e / (e - pi)
Therefore, the series converges, and its sum is e / (e - pi).