Comparison Test problem with infinite series

TheRascalKing
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Homework Statement


I need to use the Comparison Test or the Limit Comparison Test to determine whether or not this series converges:

∑ sin(1/n^2) from 1 to ∞


Homework Equations



Limit Comparison Test: Let {An} and {Bn} be positive sequences. Assume the following limit exists:
L = lim[n→∞] An/Bn
if L>0, then ƩAn converges iff ƩBn converges.
if L = ∞ and ƩAn converges, then ƩBn converges.
if L = 0 and ƩBn converges, then ƩAn converges.

Comparison Test: Assume that there exists M > 0 such that 0 ≤ An ≤ Bn for n ≥ M.
if Ʃ[n=1 to ∞] Bn converges, then Ʃ[n=1 to ∞]An also converges.
if Ʃ[n=1 to ∞] An diverges, then Ʃ[n=1 to ∞]Bn also diverges.

The Attempt at a Solution



I've tried comparing with sin(1/n), sin(n), sin(1/n^3), sin(1/n^4), and a handful of other functions involving sin.

Sorry, I'm new to the comparison test and limit comparison test :/
 
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TheRascalKing said:

Homework Statement


I need to use the Comparison Test or the Limit Comparison Test to determine whether or not this series converges:

∑ sin(1/n^2) from 1 to ∞


Homework Equations



Limit Comparison Test: Let {An} and {Bn} be positive sequences. Assume the following limit exists:
L = lim[n→∞] An/Bn
if L>0, then ƩAn converges iff ƩBn converges.
if L = ∞ and ƩAn converges, then ƩBn converges.
if L = 0 and ƩBn converges, then ƩAn converges.

Comparison Test: Assume that there exists M > 0 such that 0 ≤ An ≤ Bn for n ≥ M.
if Ʃ[n=1 to ∞] Bn converges, then Ʃ[n=1 to ∞]An also converges.
if Ʃ[n=1 to ∞] An diverges, then Ʃ[n=1 to ∞]Bn also diverges.

The Attempt at a Solution



I've tried comparing with sin(1/n), sin(n), sin(1/n^3), sin(1/n^4), and a handful of other functions involving sin.

Sorry, I'm new to the comparison test and limit comparison test :/

Try comparing with 1/n^2. You know that converges, yes?
 
Thanks, got it now. I was using the Limit Comparison test wrong >.<
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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