Can anyone help me with Series and Convergence Problems?

Question

I have a few problems here. One I answered & the others have me stumped.

1) Determine whether or not the sequence A converges. If it converges, find its limit.

A = (sin n) / 3^n

I took the limit of A as n approaches infinity and used the squeeze theorem to get the answer A converges to 0. Is that right?

2) Determine if the infinite series converges or diverges. If it converges, find its sum.

Here is the series:
http://img.villagephotos.com/p/2006-10/1221148/Prob2.JPG

I cannot figure out what "a" is so that I can use the formula
"If |r| < 1, the series diverges to a/(1-r)." I'm assuming r = 9/10

3) Express the nth partial sum of the infinite series as a telescoping sum. Find the sum of the series if it converges.

Here is the series:
http://img.villagephotos.com/p/2006-10/1221148/Prob3.JPG

I just don't know where to start, and the textbook is very confusing without clear examples of similar problems.

Thank you for your time and help!

Answer 1

#1: Yes, that is correct.
#2: This series diverges. It is a necessary (but not sufficient) condition for an infinite series to converge that the individual terms approach 0 as n→∞. It is fairly easy to show that the individual terms approach 1 as n→∞.
#3: First note that ln ((n+1)/n) = ln (n+1) - ln n. So writing out the first few terms of this series:

(ln 2 - ln 1) + (ln 3 - ln 2) + (ln 4 - ln 3) + ...

Note that the second part of each term cancels out the first part of the preceding term. So all that needs to be considered is the first part of the last term and the last part of the first term, because those are the only parts that do not cancel. Formally:
[n=1, a]∑ln ((n+1)/n) = ln (a+1) - ln 1 = ln (a+1)

This series does not converge, since [a→∞]lim ln (a+1) = ∞.

Answer 2

mathsmanretired's answer in reality makes use of the decrease try: the words advance and initiate off beneficial, so as that they'd't have decrease 0. Strictly conversing, the ratio try (as defined on the PlanetMath link under) is unquestionably inconclusive, because of the fact the limsup of the ratios is a million. usually i'm responding because of the fact of your L'Hopital's Rule thought. you does no longer save the factorials consistent. you will possibly % to interchange them with a sufficiently differentiable function which concurs with the factorial on the integers. The Gamma function (2d link) is the function you will possibly maximum possibly use, however this is needlessly overcomplicated while an elementary ratio calculation suffices.