How to find the exact value of $\sum_{n=1}^{\infty} \frac{\sin \left(\frac{n \pi}{4}\right)}{n^2 \cdot 2^{\frac{n}{2}}} $?

Question

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 $.

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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 $.

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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) $.

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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.

Answer 1

Rewrite $S_2=\sum_{n=1}^{\infty} \frac{\sin \left(n \cdot\frac{\pi}{4}\right) \cos ^n\left(\frac{\pi}{4}\right)}{n^2} =f\left(\frac{\pi}{4}\right)$ , where

$$ f(x)=\sum_{n=1}^{\infty} \frac{\sin \left(nx\right) \cos ^n\left(x\right)}{n^2} $$

Differentiating w.r.t. $x$, we have $$ \begin{aligned} f^{\prime}(x) &= \sum_{n=1}^{\infty}\frac{n \cos (n x) \cos ^n x-n \sin (n x) \cos ^{n-1} x \sin x}{n^2}\\ & =\underbrace{\sum_{n=1}^{\infty}\frac{\cos (n x) \cos ^n x}{n}}_{g(x)}-\tan x \cdot \underbrace{\sum_{n=1}^{\infty}\frac{\sin (n x) \cos ^n x}{n}}_{h(x)} \end{aligned} $$

Grouping $g(x)$ and $h(x) $ to form a series $$ \begin{aligned} g(x)+i h(x)& =\sum_{n=1}^{\infty} \frac{e^{i n x} \cos ^n x}{n} =\sum_{n=1}^n \frac{\left(e^{i x} \cos x\right)^n}{n} \end{aligned} $$

Using the identity for $|t|<1$, \begin{equation} -\ln (1-t)=\sum_{n=1}^{\infty} \frac{t^n}{n}, \end{equation} we have $$\begin{aligned}g(x)+ih(x)&= -\ln \left(1-e^{i x} \cos x\right)\\& =-\ln [\sin x(\sin x-i \cos x)] \\&=-\ln (\sin x)+\left(\frac{\pi}{2}-x\right) i \end{aligned}$$ Comparing real and imaginary pars yields $$g(x)=-\ln (\sin x)\text { and } h(x)=\frac{\pi}{2}-x $$ Putting back gives $$ f^{\prime} (x)=-\ln (\sin x)-\left(\frac{\pi}{2}-x\right) \tan x$$

Integrating $f^{\prime} (x)$ from $0$ to $x$, we have $$ f(x)-f(0)=-\int_0^x \ln (\sin u) d t-\int_0^x\left(\frac{\pi}{2}-u\right) \tan u d x $$ $$ \begin{aligned} f(x)= & -x \ln (\sin x)+\int_0^x \frac{u}{\tan u} d u+\frac{\pi}{2} \ln(\cos x)+\int_0^x u \tan u d u \end{aligned} $$ $$ \begin{aligned} \int_0^x \frac{u}{\tan u} d x+\int_0^x u \tan u d = & \int_0^x \frac{u \sec ^2 u}{\tan u} d u \\ = & \int_0^x u d(\ln (\tan u)) \\ = & {[u \ln (\tan u)]_0^x \int_0^x \ln (\tan u) d u } \\ = & x \ln (\tan x)-\int_0^x \ln (\tan u) d u \end{aligned} $$

$$\boxed{\sum_{n=1}^{\infty} \frac{\sin \left(nx\right) \cos ^n\left(x\right)}{n^2} = -x \ln (\sin x)+\frac{\pi}{2} \ln(\cos x)+ x \ln (\tan x)-\int_0^x \ln (\tan u) d u\;}$$ Now go back to our series,

$$\begin{aligned} S_2&=\sum_{n=1}^{\infty} \frac{\sin \left(\frac{n \pi}{4}\right)}{n^2 \cdot 2^{\frac{n}{2}}}\\&=-\frac{\pi}{4} \ln \left(\frac{1}{\sqrt{2}}\right)+\frac{\pi}{2} \ln \left(\frac{1}{\sqrt{2}}\right)- \int_0^{\pi/4} \ln (\tan u) d u \\&=G-\frac{\pi}{8} \ln 2\end{aligned}$$

where the last integral equals $-G$ (negative Catalan's constant G), from the post.

Last but not least, we may say that $$\boxed{\Im\operatorname{Li}_2\left(\frac{1+i}{2}\right) = G-\frac{\pi}{8} \ln 2 }$$

Answer 2

Too long for a comment (hoping it could be of interest to you)

Long time ago, I worked the problem of $$S_p=\sum_{n=1}^{p} \frac{\sin \left(n\frac{ \pi}{4}\right)}{n^2 \,\, 2^{\frac{n}{2}}}$$ which write (using as many $2$'s as I could) $$2^{15}\,S_p=2^{14}\,\text{Li}_4\left(\frac{1}{2}\right)-2^{6}\,\text{Li}_4\left(\frac{1}{2^4}\right)+2^4\,\Phi \left(\frac{1}{2^4},2^2,\frac{1}{2}\right)+$$ $$2^{2-4 \left\lfloor \frac{p}{2^2}\right\rfloor } \Phi \left(\frac{1}{2^4},2^2,\left\lfloor \frac{p}{2^2}\right\rfloor +1\right)-2^{13-p} \,\Phi \left(\frac{1}{2},2 ^2,p+1\right)-$$ $$2^{2^2 \left\lceil \frac{2-p}{2 ^2}\right\rceil }\Phi \left(\frac{1}{2^4},2^2,\left\lfloor \frac{p-2}{2^2}\right\rfloor +\frac{3}{2}\right)$$ where $\Phi(.)$ is the Lerch transcendent function.

What is doing the stranger $3$ here ?

The summation converges quite fast $$S_{10}- \left(C-\frac{1}{8} \pi \log (2)\right)=-3.90\times 10^{-5}$$ $$S_{100}- \left(C-\frac{1}{8} \pi \log (2)\right)=5.55\times 10^{-17}$$

Answer 3

Dilogarithm Function

Using the identity 25.12.3, we have $$ Li_2(1-z)+Li_2\left(1-z^{-1}\right)=-\frac{1}{2} \ln ^2(z) $$ Putting $z=\frac{1+i}{2}$ into it yields $$ \begin{aligned} & L_2\left(\frac{1+i}{2}\right)+L_2(-i)=-\frac{1}{2} \ln ^2\left(\frac{1-i}{2}\right) \\ \Rightarrow \quad & L_2\left(\frac{1+i}{2}\right)-\frac{\pi^2}{48}-i G=\frac{\pi^2}{32}-\frac{\ln ^2 2}{8}-\frac{\pi}{8} i \ln 2 \\ \Rightarrow \quad & L_2\left(\frac{1+i}{2}\right)=\frac{5 \pi^2}{96}-\frac{\ln ^2 2}{8}+\left(G-\frac{\pi}{8} \ln 2\right) i \end{aligned} $$ where $G$ is the Catalan’s constant. Hence $$\sum_{n=1}^{\infty} \frac{\sin \left(\frac{n \pi}{4}\right)}{n^2 \cdot 2^{\frac{n}{2}}} =G-\frac{\pi}{8} \ln 2$$

As a bonus, $$\sum_{n=1}^{\infty} \frac{\cos \left(\frac{n \pi}{4}\right)}{n^2 \cdot 2^{\frac{n}{2}}}=\frac{5 \pi^2}{96}-\frac{\ln ^2 2}{8}$$