Once I met the identity
$$ \boxed{S_0=\sum_{n=1}^{\infty} \frac{\sin \left(\frac{n \pi}{4}\right)}{2^{\frac{n}{2}}}=1}, $$
I first tried to prove it by $e^{xi}=\cos x+i\sin x$.
$$ \begin{aligned} \sum_{n=1}^{\infty} \frac{\sin \left(\frac{n \pi}{4}\right)}{2^{\frac{n}{2}}} & =\Im\left[\sum_{n=1}^{\infty}\left(\frac{e^{\frac{\pi}{4} i}}{\sqrt{2}}\right)^n\right]\\& =\Im\left(\frac{e^{\frac{\pi}{4} i} / \sqrt{2}}{1-e^{\frac{\pi}{4} i} / \sqrt{2}}\right)\\&=\Im\left(\frac{1+i}{1-i}\right)\\&=1 \end{aligned} $$
and consequently $C_0=\sum_{n=1}^{\infty} \frac{\cos \left(\frac{n \pi}{4}\right)}{ 2^{\frac{n}{2}}}=0 $.
Similarly, I try further with
$$ S_1=\sum_{n=1}^{\infty} \frac{\sin \left(\frac{n \pi}{4}\right)}{n \cdot 2^{\frac{n}{2}}}= \begin{equation} \Im\left[\sum_{n=1}^{\infty} \frac{\left(e^{\frac{\pi}{4} i} / \sqrt{2}\right)^n}{n}\right] \end{equation} $$ Inspired by the post, I used the Taylor expansion of $\ln(1-t)$ for $|t|<1$.
\begin{equation} -\ln (1-t)=\sum_{n=1}^{\infty} \frac{t^n}{n} \end{equation}
$$ \begin{aligned}\sum_{i=1}^{\infty} \frac{\left(e^{\frac{\pi}{4} i}/ \sqrt{2}\right)^n}{n} & =-\ln \left(1-\frac{e^{\frac{\pi}{4} i}}{\sqrt{2}}\right) \\ & =\frac{1}{2}\ln 2+ \frac{\pi}{4} i \end{aligned} $$ So we got
$$\boxed{S_1=\sum_{n=1}^{\infty} \frac{\sin \left(\frac{n \pi}{4}\right)}{n \cdot 2^{\frac{n}{2}}}=\frac{\pi}{4} } $$ and consequently $C_1=\sum_{n=1}^{\infty} \frac{\cos \left(\frac{n \pi}{4}\right)}{n \cdot 2^{\frac{n}{2}}}=\frac{1}{2}\ln 2 $.
Both results are so nice that I then kept going with $S_2$ using the dilogarithm function $\operatorname{Li}_2$
$$\boxed{\begin{aligned}S_2&=\sum_{n=1}^{\infty} \frac{\sin \left(\frac{n \pi}{4}\right)}{n^2 \cdot 2^{\frac{n}{2}}}= \Im\left[\sum_{n=1}^{\infty} \frac{\left(e^{\frac{\pi}{4} i} / \sqrt{2}\right)^n}{n^2}\right]\\& = \Im\left[\sum_{n=1}^{\infty} \frac{1}{n^2}\left(\frac{1+i}{2}\right)^n\right] =\Im \operatorname{Li}_2\left(\frac{1+i}{2}\right)\end{aligned} }$$ and consequently $C_2= \sum_{n=1}^{\infty} \frac{\cos \left(\frac{n \pi}{4}\right)}{n^2 \cdot 2^{\frac{n}{2}}}=\Re \operatorname{Li}_2\left(\frac{1+i}{2}\right) $.
My question: How to find the exact value of $S_2$ or $\operatorname{Li}_2\left(\frac{1+i}{2}\right) $?
Your comments and solutions are highly appreciated.