Question

Σ ((cos x)^2)/(n^2 + 1) from 1 to infinity Σ ((cos x)^2)/(n^2 + 1) from 1 to infinity @Mathematics

Answer 1

\[\sum_{1}^{\infty} \cos ^{2} x /n ^{2} + 1\]

Answer 2

Are you asking what that sum is, or whether or not it converges?

Answer 3

From Mathematica 8.04:\[\sum _{n=1}^{\text{Infinity}} \frac{\cos ^2(x)}{n^2+1}\text{=}\frac{1}{2} (\pi \coth (\pi )-1) \cos ^2(x)\]

Answer 4

The actual sum is pretty gross, so I'll assume you mean to ask if it converges. The comparison test says that if we have two series a and b where for all n,
\[a_n \geq b_n \]
and a converges, then b must also converge. We can use this by noting that, since
\[ cos^2(x) \leq 1 \]
\[\frac{cos^2(x)}{n^2+1} \leq \frac{1}{n^2+1} \]
and furthermore, since
\[ n^2+1 > n^2 \]
we have that
\[ \frac{cos^2(x)}{n^2+1} \leq \frac{1}{n^2+1} < \frac{1}{n^2} \]
and since the series
\[\sum\frac{1}{n^2}\]
converges, then our series converges too.

Answer 5

awesome thank you! but how do you know that 1/n^2 converges?

Answer 6

That is a pretty common series called a p-series. It turns out that
\[\sum\frac{1}{n^p} \]
converges for all p greater than or equal to 1. You can prove this via the integral test, which says that if you have some sequence a, and you convert that sequence to a function, and that function's integral converges, then the sequence converges too. For example,
\[ a_n = \frac{1}{n^p} \]
let
\[ f(x) = \frac{1}{x^p} \]
assuming
\[p \geq 1\]
we have that
\[ \int_1^{\infty} \frac{1}{x^p} dx = -(p-1)\frac{1}{x^{p-1}}|^{x=\infty}_{x=1} = p-1\]
thus the integral converges, and the corresponding sequence also converges, if
\[p \geq 1\]

Answer 7

http://www.wolframalpha.com/input/?_=1320267044545&i=sum+from+1+to+infinity+cos^2(x)%2f(n^2+%2b+1)&fp=1&incTime=true