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:
converges, the series:
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`
`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.