Evaluate $\sum_{n=2}^{\infty}\frac{n^4+3n^2+10n+10}{2^n(n^4+4)}$ Evaluate $\sum_{n=2}^{\infty}\frac{n^4+3n^2+10n+10}{2^n(n^4+4)}$
I am not able to convert it into telescopic function.

So the series
$$
\sum_{n=2}^{\infty}a_n\tag{1}
$$
has to be written in the form
$$
\sum_{n=2}^{\infty} (b_{n} - b_{n-1});\tag{2}
$$
therefore
$$
b_1 = \sum_{n=2}^{\infty} a_n, \\
b_2 = \sum_{n=3}^{\infty} a_n, \\
b_3 = \sum_{n=4}^{\infty} a_n, \\
\cdots \tag{3}
$$
If search $b_n$ in the form $$b_n = \dfrac{1}{2^n}\dfrac{P_4(n)}{Q_4(n)},$$
where $P_4(n), Q_4(n) $ are polynomials of $4$th degree; then one can figure out that
$$
b_n = \dfrac{1}{2^n}\left(1 + \dfrac{3n^2+4n+5}{n^4+2n^3+3n^2+2n+2}\right) \\ 
=  \dfrac{1}{2^n}\left(1 + \dfrac{3n^2+4n+5}{(n^2+1)\Bigl((n+1)^2+1\Bigr)}\right).
$$
Then $$
b_n - b_{n-1} = (weird \;expression) = \dfrac{n^4+3n^2+10n+10}{2^n(n^3+4)} = a_n.
$$
So
$$
\sum_{n=2}^{\infty} a_n = b_{1} = \dfrac{1}{2}\cdot \dfrac{22}{10}=\dfrac{11}{10}.
$$
WolframAlpha checking link .
