Sunday

November 23, 2014

November 23, 2014

Posted by **Anonymous** on Friday, February 22, 2008 at 9:59pm.

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)

----------------------------

1/2 + 2/3 + 3/4 + 4/5 + ... + n/(n+1)+...

even number question. don't know the answer

---------------------------

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)

- Calculus (#2 of 3) -
**drwls**, Saturday, February 23, 2008 at 7:12amThe 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.

- Calculus (#1 and #3 of 3) -
**drwls**, Saturday, February 23, 2008 at 7:19amI 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

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