Posted by Anonymous on Friday, February 22, 2008 at 9:59pm.
The problem with these two questions is that I cannot determine the a and r. The 3rd questionI 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, 2sqrt(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/(epi)
work:
1<e/pi<1 Therefore, converges
a=1, r=e/pi
S= 1/(1e/pi)
=1/(pi/pie/pi)
=1/[(pie)/pi]
=pi/(pie) (not the same as the answer)

Calculus (#2 of 3)  drwls, Saturday, February 23, 2008 at 7:12am
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/(epi) 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 7:19am
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
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