How to approach $\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}$?

Question

@User mentioned in the comments that

$$\sum _{n=1}^{\infty } \frac{16^n}{n^3 \binom{2 n}{n}^2}=8\pi\text{G}-14 \zeta (3)\tag1$$

$$\small{\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}=64 \pi \Im(\text{Li}_3(1+i))+64 \text{Li}_4\left(\frac{1}{2}\right)-233 \zeta(4)-40 \ln ^2(2)\zeta(2)+\frac{8}{3}\ln ^4(2)}\tag2$$

I was able to prove $(1)$ but had some difficulty proving $(2)$. Any idea?

I am going to show my proof of $(1)$ hoping it helps you prove $(2)$:

We showed in this question that

$$\sum_{n=1}^\infty\frac{4^ny^n}{n^2{2n\choose n}}=2\int_0^y \frac{\arcsin \sqrt{x}}{\sqrt{x}\sqrt{1-x}}dx$$

multiply both sides by $\frac{1}{y\sqrt{1-y}}$ then $\int_0^1$ with respect to $y$ and use $\int_0^1\frac{y^{n-1}}{\sqrt{1-y}}dy=\frac{4^n}{n{2n\choose n}}$ we obtain

$$\sum _{n=1}^{\infty } \frac{16^n}{n^3 \binom{2 n}{n}^2}=2\int_0^1\int_0^y \frac{\arcsin \sqrt{x}}{y\sqrt{x}\sqrt{1-x}\sqrt{1-y}}dxdy$$

$$=2\int_0^1\frac{\arcsin\sqrt{x}}{\sqrt{x}\sqrt{1-x}}\left(\int_x^1\frac{dy}{y\sqrt{1-y}}\right)dx$$

$$=2\int_0^1\frac{\arcsin\sqrt{x}}{\sqrt{x}\sqrt{1-x}}\left(2\ln(1+\sqrt{1-x})-\ln x\right)dx$$

$$\overset{\sqrt{x}=\sin \theta}{=}8\int_0^{\pi/2}x\ln(1+\cos x)dx-8\int_0^{\pi/2}x\ln(\sin x)dx$$

$$=8\int_0^{\pi/2}x\ln(2\cos^2\frac x2)dx-8\int_0^{\pi/2}x\ln(\sin x)dx$$

$$=32\int_0^{\pi/4}x\ln(2\cos^2x)dx-8\int_0^{\pi/2}x\ln(\sin x)dx$$

$$=32\underbrace{\int_0^{\pi/4}x\ln(2)dx}_{\frac3{16}\ln(2)\zeta(2)}+64\underbrace{\int_0^{\pi/4}x\ln(\cos x)dx}_{\frac{\pi}{8}\text{G}-\frac3{16}\ln(2)\zeta(2)-\frac{21}{128}\zeta(3)}-8\underbrace{\int_0^{\pi/2}x\ln(\sin x)dx}_{\frac7{16}\zeta(3)-\frac34\ln(2)\zeta(2)}$$

$$=8\pi\text{G}-14 \zeta (3)$$

The last two integrals follow from using the Fourier series of $\ln(\cos x)$ and $\ln(\sin x)$.

All approaches are appreciated. Thank you.

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Addendum: Here is an easier way to prove $(1)$:

We have

$$\arcsin^2(x)=\frac12\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2{2n\choose n}}$$

or

$$\sum_{n=1}^\infty\frac{4^nx^n}{n^2{2n\choose n}}=2\arcsin^2(\sqrt{x})$$

Divide both sides by $x\sqrt{1-x}$ then $\int_0^1$ and use $\int_0^1\frac{x^{n-1}}{\sqrt{1-x}}dx=\frac{4^n}{n{2n\choose n}}$ we have

$$\sum_{n=1}^\infty\frac{16^n}{n^3{2n\choose n}^2}=2\int_0^1\frac{\arcsin^2(\sqrt{x})}{x\sqrt{1-x}}dx$$

$$\overset{\sqrt{x}=\sin x}{=}4\int_0^{\pi/2}x^2 \csc(x)dx$$

$$\overset{IBP}{=}-8\int_0^{\pi/4} x\ln(\tan\frac x2)dx=8\pi\text{G}-14\zeta(3)$$

where the last result follows from the Fourier series of $\ln(\tan\frac x2)$.

Answer 1

Too long for a comment (from Cornel)

Well, the elementary tools presented by OP are enough to immediately get a reduction to single integrals by simple integrations by parts and changing of integration order. So, the series is equal to $$\sum _{n=1}^{\infty } \frac{16^n}{\displaystyle n^4 \binom{2 n}{n}^2}=\int _0^1\frac{1}{z\sqrt{1-z}}\left(\int _0^z\frac{1}{y}\left(\int _0^y\frac{2 \arcsin(\sqrt{x})}{\sqrt{x (1-x)}}\textrm{d}x \right)\textrm{d}y \right)\textrm{d}z$$ $$=-32\int_0^1 \frac{\arctan^2(x)\log (x)}{x} \textrm{d}x-\frac{64}{3} \int_0^1 \arctan^3(x) \textrm{d}x-\frac{64}{3} \int_0^1 \arctan^3(x)\log (x)\textrm{d}x,$$

and the desired result follows from using that

$$\int_0^{1} \frac{\arctan(x)^2\log (x)}{x} \textrm{d}x$$ $$=\operatorname{Li}_4\left(\frac{1}{2}\right)+\frac{1}{24}\log ^4(2)+\frac{7}{8}\log (2)\zeta (3) -\frac{151 }{11520}\pi ^4-\frac{1}{24}\log ^2(2)\pi ^2,$$ which requires some special techniques. For example, user Song has already posted on site a solution where contour integration is cleverly exploited, but also other clever ways are possible.

Then,

$$\int_0^1 \arctan^3(x) \textrm{d}x=\frac{\pi ^3}{64}+\frac{3}{32} \pi ^2 \log (2)-\frac{3 }{4}\pi G+\frac{63 }{64}\zeta(3),$$

which is trivial (variable change and Fourier series).

Next,

$$ \int_0^1 \arctan^3(x)\log (x)\textrm{d}x$$ $$=\frac{3 }{4}\pi G-\frac{3}{32} \log (2)\pi ^2+\frac{3}{8} \log ^2(2) \pi ^2-\frac{\pi ^3}{64}+\frac{361 }{2560}\pi ^4-\frac{63 }{64}\zeta (3)-\frac{21}{16} \log (2)\zeta (3) -\frac{3}{16}\log ^4(2)-3 \pi \Im\{\text{Li}_3(1+i)\}-\frac{9 }{2}\operatorname{Li}_4\left(\frac{1}{2}\right),$$ which combine Fourier series and the method of Random Variable in this post Looking for closed-forms of $\int_0^{\pi/4}\ln^2(\sin x)\,dx$ and $\int_0^{\pi/4}\ln^2(\cos x)\,dx$. The fourier series in the book, (Almost) Impossible Integrals, Sums, and Series, page $243$, eq. $3.281$, may also be found extremely useful after the integral tranformation into a trigonometric one. Furthermore, good to know that instead of Random Variable's way where necessary we can try to adjust and use the strategy in this post, https://math.stackexchange.com/q/3798026.

A first note: By similar means, one can calculate the version, $$\displaystyle \sum _{n=1}^{\infty } \frac{16^n}{\displaystyle n^5 \binom{2 n}{n}^2}.$$

A second note: Most apparently advanced integrals and series flying around the site this period of time are easily manageable mostly by simple techniques. For example, one can calculate advanced nontrivial harmonic series of weights, $8$, $9$, $10$, $11$, $12$ by only combining and using elementary identities with harmonic numbers, nothing advanced is necessary. Surely, advanced methods are embraced and appreciated as well.

Answer 2

Since

$$\frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^\infty\frac{(2x)^{2n-1}}{n{2n\choose n}}$$

we can write

$$\sum_{n=1}^\infty\frac{4^nx^{n}}{n{2n\choose n}}=\frac{2\sqrt{x}\arcsin \sqrt{x}}{\sqrt{1-x}}$$

Multiply both sides by $-\frac{\ln x}{x}$ then $\int_0^y$ and use the fact that $\int_0^y - x^{n-1}\ln xdx=\frac{1}{n^2}y^n-\frac{\ln y}{n}y^n$

$$\sum_{n=1}^\infty\frac{4^ny^n}{n^3{2n\choose n}}-\ln y\sum_{n=1}^\infty\frac{4^ny^n}{n^2{2n\choose n}}=-\int_0^y \frac{2\ln x\arcsin \sqrt{x}}{\sqrt{x}\sqrt{1-x}}dx$$

Next multiply both sides by $\frac{1}{y\sqrt{1-y}}$ then $\int_0^1$ we get

$$\sum_{n=1}^\infty\frac{4^n}{n^3{2n\choose n}}\left(\int_0^1\frac{y^{n-1}}{\sqrt{1-y}}dy\right)-\int_0^y\frac{\ln y}{y\sqrt{1-y}}\left(\sum_{n=1}^\infty\frac{(2\sqrt{y})^{2n}}{n^2{2n\choose n}}\right)dx$$ $$=-\int_0^1\int_0^y \frac{2\ln x\arcsin \sqrt{x}}{y\sqrt{x}\sqrt{1-x}\sqrt{1-y}}dxdy=-\int_0^1 \frac{2\ln x\arcsin \sqrt{x}}{\sqrt{x}\sqrt{1-x}}\left(\int_x^1\frac{dy}{y\sqrt{1-y}}\right)dx$$

$$=-\int_0^1 \frac{2\ln x\arcsin \sqrt{x}}{\sqrt{x}\sqrt{1-x}}\left(2\ln(1+\sqrt{1-x})-\ln x\right)dx$$

$$\overset{\sqrt{x}=\sin\theta}{=}16\int_0^{\pi/2}x\ln(\sin x)\ln\left(\frac{\sin x}{1+\cos x}\right)dx$$

$$=16\int_0^{\pi/2}x\ln(\sin x)\ln\left(\tan(\frac x2)\right)dx$$

$$\overset{x\to 2x}{=}64\int_0^{\pi/4}x\ln(\sin(2x))\ln\left(\tan x\right)dx$$

$$=64\int_0^{\pi/4}x[\ln(2)+\ln(\sin x)+\ln(\cos x)][\ln(\sin x)-\ln(\cos x)]dx$$

$$=64\ln(2)\int_0^{\pi/4}x\ln(\tan x)dx+64\int_0^{\pi/4}x\ln^2(\sin x)dx-64\int_0^{\pi/4}x\ln^2(\cos x)dx$$

For the LHS, use $\int_0^1\frac{y^{n-1}}{\sqrt{1-y}}dy=\frac{4^n}{n{2n\choose n}}$ and $\sum_{n=1}^\infty\frac{(2\sqrt{y})^{2n}}{n^2{2n\choose n}}=2\arcsin^2(\sqrt{y})$ we get

$$\text{LHS}=\sum_{n=1}^\infty\frac{16^n}{n^4{2n\choose n}^2}-2\int_0^1\frac{\ln y\arcsin^2(\sqrt{y})}{y\sqrt{1-y}}dy$$ $$\overset{\sqrt{y}=\sin \theta}{=}\sum_{n=1}^\infty\frac{16^n}{n^4{2n\choose n}^2}-8\int_0^{\pi/2} x^2\csc x\ln(\sin x)dx$$

Therefore

$$\sum_{n=1}^\infty\frac{16^n}{n^4{2n\choose n}^2}=64\ln(2)\int_0^{\pi/4}x\ln(\tan x)dx-64\int_0^{\pi/4}x\ln^2(\cos x)dx$$ $$+64\int_0^{\pi/4}x\ln^2(\sin x)dx+8\int_0^{\pi/2} x^2\csc x\ln(\sin x)dx\tag1$$

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The first integral can be done via Fourier series:

$$\int_0^{\pi/4} x\ln(\tan x)dx=\frac{7}{16}\zeta(3)-\frac{\pi}{4}\text{G}\tag2$$

The second integral:

$$\int_0^{\pi/4}x\ln^2(\cos x)dx=\int_0^{\pi/2}x\ln^2(\cos x)dx-\underbrace{\int_{\pi/4}^{\pi/2}x\ln^2(\cos x)dx}_{x\to \pi/2-x}$$

$$=\int_0^{\pi/2}x\ln^2(\cos x)dx-\int_{\pi/4}^{\pi/2}(\frac{\pi}{2}-x)\ln^2(\sin x)dx$$

$$=\int_0^{\pi/2}x\ln^2(\cos x)dx-\frac{\pi}{2}\int_0^{\pi/4}\ln^2(\sin x)dx+\int_0^{\pi/4}x\ln^2(\sin x)dx$$

Plugging this result along with $(2)$ in $(1)$, the integral $\int_0^{\pi/4}x\ln^2(\sin x)dx$ nicely cancels out getting:

$$\sum_{n=1}^\infty\frac{16^n}{n^4{2n\choose n}^2}=28\ln(2)\zeta(3)-16\pi\ln(2)\text{G}-64\int_0^{\pi/2}x\ln^2(\cos x)dx$$ $$+32\pi\int_0^{\pi/4}\ln^2(\sin x)dx+8\int_0^{\pi/2} x^2\csc x\ln(\sin x)dx$$

Lets manipulate the first integral using the same trick $x\to \pi/2-x$:

$$\int_0^{\pi/2}x\ln^2(\cos x)dx=\int_0^{\pi/2}(\frac{\pi}{2}-x)\ln^2(\sin x)dx$$

$$=\frac{\pi}{2}\int_0^{\pi/2}\ln^2(\cos x)dx-\int_0^{\pi/2}x\ln^2(\sin x)dx$$

By Beta function we have

$$\frac{\pi}{2}\int_0^{\pi/2}\ln^2(\cos x)dx=\frac{15}{8}\zeta(4)+\frac32\ln^2(2)\zeta(2)$$

and our sum boils down to

$$\sum_{n=1}^\infty\frac{16^n}{n^4{2n\choose n}^2}=28\ln(2)\zeta(3)-16\pi\ln(2)\text{G}-120\zeta(4)-96\ln^2(2)\zeta(2)$$ $$+64\underbrace{\int_0^{\pi/2}x\ln^2(\sin x)dx}_{\mathcal{\Large{I_1}}}+32\pi\underbrace{\int_0^{\pi/4}\ln^2(\sin x)dx}_{\mathcal{\Large{I_2}}}+8\underbrace{\int_0^{\pi/2}x\csc x\ln(\sin x)dx}_{\mathcal{\Large{I_3}}}$$

$\mathcal{I}_1$ is calculated here:

$$\int_0^{\pi/2} x\ln^2(\sin x)\textrm{d}x=\frac{1}{2}\ln^2(2)\zeta(2)-\frac{19}{32}\zeta(4)+\frac{1}{24}\ln^4(2)+\operatorname{Li}_4\left(\frac{1}{2}\right)$$

$\mathcal{I}_2$ is calculated here

$$\int_{0}^{\pi /4} \ln^{2}(\sin x) \ dx = \frac{\pi^{3}}{192} + G\frac{ \ln(2)}{2} + \frac{3 \pi}{16} \ln^{2}(2) + \text{Im} \ \text{Li}_{3}(1+i).$$

$\mathcal{I}_3$ is calculated here

$$\int_0^{\pi/2} \frac{x^2 \ln(\sin x)}{\sin (x)} dx=-4 \pi \Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}-\frac{7}{2} \zeta (3) \ln (2)+\frac{135}{16}\zeta(4)+\frac{3}{4} \zeta(2) \ln ^2(2)$$ $$=4\pi\Im\{\text{Li}_3(1+i)\}-\frac{45}{4}\zeta(4)-\frac72\ln(2)\zeta(3)-\frac32\ln^2(2)\zeta(2)$$

The last result follows from using

$$\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}=\frac{7\pi^3}{128}+\frac{3\pi}{32}\ln^2(2)-\Im\{\text{Li}_3(1+i)\}$$

Collecting the three integrals we finally get

$$\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}=64 \pi \Im\{\text{Li}_3(1+i)\}+64 \text{Li}_4\left(\frac{1}{2}\right)-233 \zeta(4)-40 \ln ^2(2)\zeta(2)+\frac{8}{3}\ln ^4(2)$$

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Thanks to Cornel for the hint $x\to \pi/2-x$ which simplifies $\int_0^{\pi/2}x\ln^2(\cos x)dx$ to known integrals.