Sum $\sum_{n=2}^{\infty} \frac{n^4+3n^2+10n+10}{2^n(n^4+4)}$ I want to evaluate the sum $$\large\sum_{n=2}^{\infty} \frac{n^4+3n^2+10n+10}{2^n(n^4+4)}.$$  I did partial fraction decomposition to get $$\frac{1}{2^n}\left(\frac{-1}{n^2+2n+2}+\frac{4}{n^2-2n+2}+1\right)$$ I am absolutely stuck after this.

You are almost there. 
So once you have obtained $$\frac{1}{2^n}\left(\frac{-1}{n^2+2n+2}+\frac{4}{n^2-2n+2}+1\right)$$ then observe the following:
$n^2+2n+2=(n+1)^2+1$ and $n^2-2n+2=(n-1)^2+1$. Which means the given sum becomes 
$$\sum\limits_{n=2}^\infty \left[\frac{-1}{2^n\{(n+1)^2+1\}}+\frac{1}{2^{n-2}\{(n-1)^2+1\}}   \right] +\sum\limits_{n=2}^\infty \frac{1}{2^n}=:\sum\limits_{n=2}^\infty [u_n-u_{n+2}]+\frac 12  $$ where $u_n:=\frac{1}{2^{n-2}\{(n-1)^2+1\}}$.
If you calculate $\sum\limits_{n=2}^m[u_n-n_{n+2}]$, you will find $u_2+u_3-(u_{m+1}+u_{m+2})$. hence taking $m\rightarrow \infty$ we shall get 
$$\sum\limits_{n=2}^\infty [u_n-u_{n+2}]=u_2+u_3=\frac{1}{2}+\frac{1}{10}.$$
Hence the final sum is $2\times \frac 12+\frac{1}{10}=\cdots.$
