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

ishpiro | Certified Educator

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.

oldnick | Student

`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.

pramodpandey | Student

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.