Question

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)

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

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)

Answer 1

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.

Answer 2

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