Answer
The series is convergent and the sum is $cos(1)-1$.
Work Step by Step
$\Sigma^{\infty}_{n=1} (cos\frac{1}{n^{2}}-cos\frac{1}{(n+1)^{2}})$
$S_{n}=\Sigma^{n}_{k=1} (cos\frac{1}{n^{2}}-cos\frac{1}{(n+1)^{2}})$
$=(cos\frac{1}{1^{2}}-cos\frac{1}{(1+1)^{2}})+(cos\frac{1}{2^{2}}-cos\frac{1}{(2+1)^{2}})+(cos\frac{1}{3^{2}}-cos\frac{1}{(3+1)^{2}})+...+(cos\frac{1}{n^{2}}-cos\frac{1}{(n+1)^{2}})$
$=cos(1)-cos\frac{1}{(n+1)^{2}}$
$\lim\limits_{n \to \infty}S_{n} = \lim\limits_{n \to \infty} (cos(1)-cos\frac{1}{(n+1)^{2}})$
$=cos(1) - cos(\frac{1}{\infty})$
$=cos(1) -cos(0)$
$=cos(1)-1$
Therefore the series is convergent and the sum is $cos(1)-1$
