Evaluating the sequence I am currently working on the problem.
Find $$\sum_{n=1}^{\infty} \frac{x_{n}}{n+2}$$ when $x_{n+2}=x_{n+1}-\frac{1}{2}x_{n}$ with $x_{0}=2$, $x_{1}=1$.
I was able to find $x_{n}$ to be $x_{n}=(\frac{1+i}{2})^n + (\frac{1-i}{2})^n$ which gives:
$$\sum_{n=1}^{\infty} \frac{(\frac{1+i}{2})^n + (\frac{1-i}{2})^n}{n+2}$$
I was able to make a shift of the index, $m=n+2$, which gives:
$$\sum_{m=3}^{\infty} \frac{(\frac{1+i}{2})^{m-2} + (\frac{1-i}{2})^{m-2}}{m}$$
Some simplification and substitution I was able to get:
$$4 \sum_{m=3}^{\infty} \frac{(\sqrt{2})^m}{m(2)^m} \sin(\frac{m\pi}{4})$$
I have tried to look at the partial sums but that did not help that much. I am not sure on where to go from here. If anyone can offer a hint to guide me, that would be extremely helpful. Thank you for looking.
Also, during my simplification I found that:
$$-2i \sum_{m=3}^{\infty} \frac{1}{m(2)^m} [(1+i)^{m} - (1-i)^{m}]$$
Looking at the expansion of $(1+i)^{m} - (1-i)^{m}$, the real parts cancel out leaving only the imaginary terms. I am not sure if that helps at all.
 A: Hint. An approach. One may recall that,
$$
-\log(1-z)=\sum_{n=1}^{\infty} \frac{z^n}n, \quad |z|<1,\tag1
$$ where $\displaystyle  \log (z)$ denotes the principal value of the logarithm defined by 
 \begin{align} 
  \displaystyle \log (z)  = \ln |z| + i \: \mathrm{arg}z, \quad -\pi <\mathrm{arg} z \leq \pi,\quad z \neq 0. \nonumber
\end{align}
By applying $(1)$ we obtain,
$$
\begin{align}
\sum_{n=1}^{\infty} \frac{\big(\frac{1+i}2\big)^n+\left(\frac{1-i}2\right)^n}{n+2}&=2\:\Re \sum_{n=1}^{\infty} \frac{\big(\frac{1+i}2\big)^n}{n+2}
\\&=2\:\Re\left( 2i\log\left(\frac{1-i}2 \right)-\frac32+i\right)
\\&=\Re\left( \pi-3 +i\frac{}{} (2-2 \ln 2)\right)
\\&=\pi-3,
\end{align}
$$ where we have used  $$\displaystyle 2\:\Re\left( 2i\log\left(\frac{1-i}2 \right)\right)=2\:\Re\left(2i\log\left(\frac{\sqrt{2}}2e^{\large-i\frac{\pi}4} \right)\right)=\pi.$$
Finally

$$
\sum_{n=1}^{\infty} \frac{x_{n}}{n+2}=\pi-3.
$$

A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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Indeed, you don't need the $\ds{x_{n}\ \mbox{v.s.}\ n}$ explicit expression. You can 'jump' right away into the generating function.


With $\ds{\quad x_{0} =2\,,\quad x_{1} = 1}$:

\begin{align}
0 & = \sum_{n = 0}^{\infty}\pars{2\,x_{n + 2} - 2\,x_{n +1} + x_{n}}z^{n + 3} =
2\sum_{n = 2}^{\infty}x_{n}\,z^{n + 1} - 2z\sum_{n = 1}^{\infty}x_{n}\,z^{n + 1} +
z^{2}\sum_{n = 0}^{\infty}x_{n}\,z^{n + 1}
\\[4mm] & =
2\pars{-x_{1}\,z^{2} + \sum_{n = 1}^{\infty}x_{n}\,z^{n + 1}} -
2z\sum_{n = 1}^{\infty}x_{n}\,z^{n + 1} + z^{2}\pars{x_{0}\,z +
\sum_{n = 1}^{\infty}x_{n}\,z^{n + 1}}
\\[4mm] & =
2z^{3} - 2z^{2} + \pars{2 - 2z + z^{2}}\sum_{n = 1}^{\infty}x_{n}\,z^{n + 1}
\end{align}

$$
\imp\quad\sum_{n = 1}^{\infty}x_{n}\,z^{n + 1} =
-2\,{z^{3} - z^{2} \over z^{2} - 2z + 2}
$$

Integrate both sides over $\ds{\pars{0,1}}$:
$$
\color{#f00}{\sum_{n = 1}^{\infty}{x_{n} \over n + 2}} =
-2\int_{0}^{1}{z^{3} - z^{2} \over z^{2} - 2z + 2}\,\dd z =
\color{#f00}{\pi - 3}
$$
