Is there a closed form for $\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot\left({2n-1\over 2n+1}\right)^2?$

Question

We have

$$\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot{2n-1\over 2n+1}=4-\pi\tag1$$

I would like to know if there exist a closed form for

$$\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot\left({2n-1\over 2n+1}\right)^2 =\,??\tag2$$

I was able to roughly estimate it as $\approx\sqrt{8+2\pi}$ but it is not the closed form.

How can we find the closd form for $(2)?$

Answer 1

Let us denote: \begin{equation} S_p := \sum\limits_{n=0}^\infty \frac{2^{n+1}}{\binom{2 n}{n}} \cdot \frac{1}{(2n+1)^p} \end{equation} then the sum in question is just equal $S_0 - 4 S_1+4 S_2$. Now we have: \begin{eqnarray} S_0 &=& \sum\limits_{n=0}^\infty 2^{n+1} \cdot \underbrace{(2n+1)}_{\left. d_\theta \theta^{2n+1} \right|_{\theta=1}} \cdot \int\limits_0^1 t^n (1-t)^n dt \\ &=& \left. d_\theta 2 \theta \int\limits_0^1 \frac{1}{1-2 \theta^2 t (1-t)} dt \right|_{\theta=1} \\ &=& \left. d_\theta \frac{4 \arctan(\frac{\theta}{\sqrt{2-\theta^2}})}{\sqrt{2-\theta^2}} \right|_{\theta=1} = 4+\pi \end{eqnarray} We compute $S_1$ in exactly the same way. We have: \begin{eqnarray} S_1 &=& \sum\limits_{n=0}^\infty 2^{n+1} \int\limits_0^1 t^n (1-t)^n dt \\ &=& 2 \int\limits_0^1 \frac{1}{1-2 t(1-t) } dt = \frac{4 \arctan(1)}{1} = \pi \end{eqnarray} Now comes a little harder task. We have: \begin{eqnarray} S_2 &=& \sum\limits_{n=0}^\infty 2^{n+1} \cdot \underbrace{\frac{1}{(2n+1)}}_{\int\limits_0^1 \theta^{2 n} dt}\cdot \int\limits_0^1 t^n (1-t)^n dt \\ &=& 2 \int\limits_0^1 \int\limits_0^1 \frac{1}{1-2 \theta^2 t (1-t)} dt d\theta \\ &=& \int\limits_0^1 \frac{4 \arctan(\frac{\theta}{\sqrt{2-\theta^2}})}{\theta \sqrt{2-\theta^2}} d \theta \\ &\underbrace{=}_{u = \theta/\sqrt{2-\theta^2}}& 2 \sqrt{2} \int\limits_0^1 \frac{\arctan(u)}{u \sqrt{1+u^2}} du \\ &\underbrace{=}_{v=\arctan(u)}&2 \sqrt{2} \int\limits_0^{\frac{\pi}{4}} \frac{v}{\sin(v)} dv \\ &=& 2 \sqrt{2} \left.\left[v \left( \log(1-e^{\imath \cdot v}) - \log(1+e^{\imath \cdot v}) \right) + \imath \left( Li_2(-e^{\imath \cdot v}) - Li_2(e^{\imath \cdot v})\right)\right]\right|_{v=0}^{v=\pi/4} \\ &=& \sqrt{2} \frac{\pi}{2} \left( \left(\log(1-e^{\frac{\imath \pi}{4}}) - \log(1+e^{\frac{\imath \pi}{4}})\right) + \imath \frac{4}{\pi} \left( Li_2(-e^{\frac{\imath \pi}{4}}) - Li_2(e^{\frac{\imath \pi}{4}})\right)+\imath \pi\right)\\ &=& \frac{1}{16} \left( \zeta(2,\frac{1}{8})+\zeta(2,\frac{3}{8})-\zeta(2,\frac{5}{8})-\zeta(2,\frac{7}{8})\right) + \frac{\pi}{\sqrt{2}} \log(-1+\sqrt{2}) \end{eqnarray} In the last line we used the following identities: \begin{eqnarray} &&\log\left[ 1- e^{\imath \pi/4}\right] - \log\left[ 1+ e^{\imath \pi/4}\right] = \log\left[\sqrt{2}-1\right] - \imath \frac{\pi}{2}\\ &&Li_2(-e^{\frac{\imath \pi}{4}}) - Li_2(e^{\frac{\imath \pi}{4}}) = -\frac{1}{32 \sqrt{2}} \left( \right. \\ &&\zeta(2,\frac{1}{8}) - \zeta(2,\frac{3}{8}) - \zeta(2,\frac{5}{8})+\zeta(2,\frac{7}{8}) + \imath \left( \zeta(2,\frac{1}{8}) + \zeta(2,\frac{3}{8}) - \zeta(2,\frac{5}{8})-\zeta(2,\frac{7}{8})\right)\left.\right) \end{eqnarray} Here we only note that for generic $p\ge 1$ we have: \begin{eqnarray} &&S_{p+1} = -4 \imath \sqrt{2} \sum\limits_{t=1}^p \sum\limits_{s=0}^{p-t} \\ && \frac{(-1)^s}{2^{p-1} (p-s-t)! s! (t-1)!} (\log(-2))^{p-s-t} 2^{s+t-1} \int\limits_1^{\exp(\imath \pi/4)} \frac{[\log(z^2-1)]^s [\log(z)]^t}{z^2-1} d z \end{eqnarray} In general we also have: \begin{eqnarray} &&S_{p+1}(x) = -\imath 8 x \sum\limits_{t=1}^p \sum\limits_{s=0}^{p-t} \frac{(-1)^s}{(p-s-t)!s! (t-1)!} \cdot \left( \imath \frac{\pi}{2}+\log(2 x)\right)^{p-s-t} \cdot \\ &&\int\limits_1^{\exp(\imath \arcsin(x))} \frac{[\log(z^2-1)]^s [\log(z)]^t}{z^2-1} dz \end{eqnarray} where \begin{equation} S_{p+1}(x) := \sum\limits_{n=0}^\infty \frac{(2 x)^{2n+2}}{\binom{2n}{n}} \cdot \frac{1}{(2n+1)^{p+1}} \end{equation} It is still not clear whether the result reduces to polylogarithms only since for the time being we are unable to find the integrals in question.

Answer 2

This is not an answer at all.

Just out of curiosity, I had (using a CAS) a look at $$S_{a,b}=\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\,\frac{(2n-1)^a}{(2n+1)^b}\qquad \qquad a\geq 1\qquad b\geq 1$$ and what it seems is that, as soon as $b>1$, the result is given as linear combinations of hypergeometric functions as Brevan Ellefsen already commented.

However, what looks to be interesting is the case $b=1$ for which very simple expressions are obtained just as in $(1)$. $$\left( \begin{array}{cc} a & S_{a,1} \\ 1 & 4-\pi \\ 2 & 5 \pi \\ 3 & 68+11 \pi \\ 4 & 512+185 \pi \\ 5 & 7300+2279 \pi \\ 6 & 116224+37085 \pi \\ 7 & 2204868+701651 \pi \\ 8 & 48073728+15302705 \pi \\ 9 & 1186130180+377556239 \pi \\ 10 & 32669570048+10399048565 \pi \end{array} \right)$$

By the way, for the approximation of you result for $a=b=2$, the simple $\frac{5000}{1323}$, $\frac{43613}{11540}$, $\frac{92226}{24403}$ correspond to very small relative errors.

Answer 3

A somewhat easier way is the following:

Let's start with

$$\sum_{n=0}^{\infty}\frac{(2x)^{2n}}{\binom{2n}{n}}=\frac{1}{1-x^2}+\frac{x\arcsin x}{(1-x^2)^{\frac{3}{2}}}$$

Now, all we need is to integrate this expression twice and to differentiate it twice with respect to $x$.

I skip these simple procedures and write down only the end result.

$$\sum_{n=0}^{\infty}\frac{(2)^{2n}(x)^{2n-2}}{\binom{2n}{n}}\left (\frac{2n-1}{2n+1}\right )^2=$$

$$=\frac{1}{x^2(1-x^2)}-\frac{4-5x^2}{x^3(1-x^2)^{\frac{3}{2}}}\arcsin x+\frac{2\arcsin^2 x}{x^4}+\frac{2}{x^3}\int_{0}^{x}\left (\frac{\arcsin t}{t}\right )^2dt$$

To get the original sum, wich we denote as $S$, we evaluate this expression at $x=\frac{1}{\sqrt{2}}$

So the final result:

$$S=\frac{\pi^2}{2}-3\pi+4+4\sqrt{2}\int_{0}^{\frac{1}{\sqrt{2}}}\left (\frac{\arcsin t}{t}\right )^2dt$$

Answer 4

This is not an answer to this question but an attempt to answer a generalized question. Our generalization consists in replacing the power two in the last term in the parentheses by a power three.Now, by starting from the (corrected--the term $(1-x^2)$ in the denominator in front of the first power of the arc sine should be raised to the power $3/2$ and not as it is now to the power one) second formula given in the answer by Martin Gales and then by multiplying the right hand side of that formula by $x^2$ and then integrating over $x$ we have derived the following formula: \begin{eqnarray} &&\sum\limits_{n=0}^\infty \frac{2^{2 n} x^{2n+1}}{\binom{2 n}{n}} \cdot \frac{(2n-1)^2}{(2n+1)^3}=\\ &&\left(4 \log(1+\sqrt{1-x^2})-4 \log(x)+ \frac{1}{\sqrt{1-x^2}}\right) \arcsin(x) + \\ &&4 \imath \left( \log\left[\frac{\left(x-\imath(1-\sqrt{1-x^2})\right)^2}{2-2 \sqrt{1-x^2}}\right] \log\left[ \frac{1-\sqrt{1-x^2}}{x}\right]-\right.\\ &&\left.Li_2[(-\imath) \frac{1-\sqrt{1-x^2}}{x} ]+Li_2[(\imath) \frac{1-\sqrt{1-x^2}}{x})]\right)+\\ &&2 \int\limits_0^x \left(\frac{\arcsin(t)}{t}\right)^2 \left(1+\log(\frac{x}{t})\right)dt =\\ &&-4 i \text{Li}_2\left(-\frac{i \left(1-\sqrt{1-x^2}\right)}{x}\right)+4 i \text{Li}_2\left(\frac{i \left(1-\sqrt{1-x^2}\right)}{x}\right)+\\ &&-8 i \text{Li}_3\left(-e^{i \sin ^{-1}(x)}\right)+8 i \text{Li}_3\left(e^{i \sin ^{-1}(x)}\right)+\\ &&\left(\text{Li}_2\left(e^{i \sin ^{-1}(x)}\right)-\text{Li}_2\left(-e^{i \sin ^{-1}(x)}\right)\right) \left(8 \sin ^{-1}(x)-4 i \log (2 i x)\right)+\\ &&\sin ^{-1}(x) \left(\frac{1}{\sqrt{1-x^2}}+4 i \log (2 i x) \cos ^{-1}\left(\frac{1}{x}\right)\right)+\\ &&4 \sin ^{-1}\left(\frac{1}{x}\right) \sin ^{-1}(x)^2+2 i \pi \log (2 i x) \cos ^{-1}(x)-14 i \zeta (3)+\\ &&8 \imath \int\limits_1^{\exp(\imath \arcsin(x))} \frac{\log(z)}{z^2-1} \log(z^2-1) dz \end{eqnarray} We will evaluate the result further and simplify it later on.