# Use the Limit Comparison Test to determine if the series is convergent: `sum_(n=0)^oo|sinn|/(n^2 +2)`

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### 3 Answers

The Limit ComparisonTest states that if `sum_(n=0)^oo a_n` and `sum_(n=0)^oo b_n`

`` are series with positive terms and `lim_(n->oo) a_n/b_n = c` , where c is a finite positive number, either both series converge or both diverge.

Suppose `sum_(n=0)^oo a_n` is the given series `sum_(n=0)^oo|sinn|/(n^2 + 2)` and let `sum_(n=0)^oo b_n=sum_(n=0)^oo 1/n^2` . This series is convergent because this is

*p*-series with *p* *= *2.

Let's take the limit of `a_n/b_n` when `n->oo:`

`` `lim_(n - > oo) a_n/b_n =lim_(n - >oo)|sin n|/(n^2 + 1) * n^2/1=lim_(n - > oo) |sin n|/(1+2/n^2)`

Since |sin n| is bounded by 0 and 1 for any n, and `2/n^2 - > 0` as `n - > oo` , this limit equals 1, which is a positive finite number.

Thus, according to the Limit Comparison Test, the given series is convergent.

`sum_(n=1)^oo |sin (n)|/(n^2+1)<=sum_(n=1)^oo 1/(n^2+1)`

for sinus function is limited in modulus by 1. Since:

`sum_(n=1)^oo 1/(n^2+1)`

converges, the series:

`sum_(n=1)^oo |sin(n)|/(n^2+1)`

converges too.

Let us define `sum_(n=1)^ooa_n ,` where

`a_n=|sin(n)|/(n^2+1)` and define an auxilary series `sum_(n=1)^oob_n ,`

`` where `b_n=1/n^2`

Note: `|sin(n)|<=1`

`a_n<b_n ` for all n.

But series `sum_(n=1)^oob_n` is convergent. so series `sum_(n=1)^ooa_n` will also converge.