Finding $\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^n 2^k {2k \choose k}}$

Question

Inspired by these two questions asking about the case n = 3 and n=4, I was wondering what is $$S =\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^n 2^k {2k \choose k}}$$ for positive integer $n \ge 3$.

For $n = 3$, the sum is $\frac{1}{4}\zeta (3)-\frac{1}{6}\ln^3(2) = \frac{1}{2}\operatorname{Li}_3\left(\frac{1}{2}\right)+\frac{1}{2}\ln(2)\operatorname{Li}_2\left(\frac{1}{2}\right)-\frac{3}{16}\zeta(3)$.

For $n = 4$, the sum is $4\operatorname{Li}_4\left(\frac12\right)-\frac72\zeta(4)+\frac{13}4\ln2\zeta(3)-\ln^22\zeta(2)+\frac5{24}\ln^42$.

Using similar work as from the two linked questions, the sum can be re-expressed as the integral $$\frac{2\cdot(-1)^{n-1}}{(n-3)!}\int_0^1\text{arcsinh}^2\left(\sqrt{\frac{x}{8}}\right)\frac{\ln^{n-3}(x)}{x}dx$$

Setting $u = \text{arcsinh}\left(\sqrt{\frac{x}{8}}\right)$, we get $$S = \frac{4\cdot(-1)^{n-1}}{(n-3)!}\underbrace{\int_0^{\frac{\ln(2)}{2}} u^2\ln^{n-3}(8\sinh^2(u))\coth(u) du}_{\large {I}}$$

$$I = \int_0^{\frac{\ln(2)}{2}}u^2\coth(u)\sum_{k=0}^{n-3}\left({n-3\choose k}\ln^{n-3-k}(8)(2\ln(\sinh(u)))^{k}\right) du$$ $$I = \sum_{k=0}^{n-3}{n-3\choose k}\ln^{n-3-k}(8)2^{k}\underbrace{\int_0^{\frac{\ln(2)}{2}}u^2\coth(u)\ln^{k}(\sinh(u))du}_{\large J}$$

Making the substitution $v = \sinh(u)$ and simplifying, we get $$J = \int_0^{\frac{1}{2\sqrt{2}}}\frac{\text{arcsinh}^2(v)\ln^k(v)}{v}dv$$

Although this may or may not help, making the substitution $w = \ln(v)$, we get $$J = \int_{-\infty}^{-\ln(2\sqrt{2})}w^k\text{arcsinh}^2(e^w)dw$$

From here, I don't know what to do to find $J$.

How can I, either through this process or a completely different one, find:

$1.$ The value of $S$ for integer $n \ge 3$?

$2.$ The value of $J$ for integer $k \ge 0$?

Answer 1

Nothing about the closed-form, but a generalization that might be helpful. (Note: The closed-form for $n = 5$ can be found in this other post.)

Consider the following function:

$$f(a,s)=\int_0^1\text{arcsinh}^2\left(\sqrt{ax}\right) x^{s-1} dx$$

It is known that:

$$\text{arcsinh} \sqrt{ax}=\text{arctanh} \frac{\sqrt{ax}}{\sqrt{1+ax}}$$

Using a generating function from this answer we can write:

$$\text{arctanh}^2 \frac{\sqrt{ax}}{\sqrt{1+ax}}= \frac{1}{2} \frac{ax}{1+ax} \sum_{m=0}^\infty \frac{H_{m+1/2}+\log 4}{m+1} \frac{a^mx^m}{(1+ax)^m}$$

Now consider the integral:

$$g_m(a,s)=\int_0^1 \frac{x^{m+s}}{(1+a x)^{m+1}} dx$$

Technically, this is a hypergeometric function, but we can express it as a more simple series, by repeatedly differentiating $g_0(a,x)$ w.r.t. $a$:

$$g_m(a,s)=\frac{(-1)^m}{m!} \sum_{k=m}^\infty \frac{(-1)^k k! a^{k-m}}{(k+s+1) (k-m)!}$$

Combining the two results, we have:

$$f(a,s)=\frac{1}{2}\sum_{m=0}^\infty (-1)^m \frac{H_{m+1/2}+\log 4}{(m+1)!} \sum_{k=m}^\infty \frac{(-1)^k k!~ a^{k+1}}{(k+s+1) (k-m)!}$$

Now consider:

$$F_n(a)=\frac{2\cdot (-1)^n}{n!}\int_0^1\text{arcsinh}^2\left(\sqrt{a x}\right)\frac{\ln^n(x)}{x}dx$$

It's clear that:

$$F_n(a)= \frac{2\cdot(-1)^n}{n!} \frac{\partial^n f(a,s)}{\partial s^n} \bigg| _{s=0}$$

Which makes for a neat double series:

$$F_n(a)=\sum_{m=0}^\infty (-1)^m \frac{H_{m+1/2}+\log 4}{m+1} \sum_{k=m}^\infty \binom{k}{m} \frac{(-1)^k a^{k+1}}{(k+1)^{n+1}}$$

$$ S=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^n 2^k {2k \choose k}}=F_{n-3} \left( \frac{1}{8} \right)$$

Note that $H_{m+1/2}+\log 4$ are rational numbers.

The original series is definitely less complicated, but it might be that the double series may offer some insight.

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As an example to judge the convergence rate of the new series, for $n=5$ we get $20$ correct digits by using the following number of terms:

$$\sum_{m=0}^{15} (-1)^m \frac{H_{m+1/2}+\log 4}{m+1} \sum_{k=m}^{15} \binom{k}{m} \frac{(-1)^k }{(k+1)^3 8^{k+1}}=0.24872280253516023269 \ldots$$

$$\sum_{k=1}^{16} \frac{(-1)^{k-1}}{k^5 2^k {2k \choose k}}=0.24872280253516023269 \ldots$$

Which means that the original sum converges faster.

But again, the generalization might be useful:

$$\sum_{k=1}^\infty \frac{(-1)^{k-1} (4a)^k}{k^n {2k \choose k}}=F_{n-3}(a)$$

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By the uniqueness of power series, we can now collect the coefficients:

$$\sum_{m=0}^\infty \frac{H_{m+1/2}+\log 4}{m+1} \sum_{k=0}^\infty \binom{m+k}{m} \frac{(-1)^k a^{m+k+1}}{(m+k+1)^{n-2}}=\sum_{l=0}^\infty \frac{(-1)^l (4a)^{l+1}}{(l+1)^n {2l+2 \choose l+1}}$$

$$k=l-m$$

$$\sum_{m=0}^\infty (-1)^m \frac{H_{m+1/2}+\log 4}{m+1} \binom{l}{m} =\frac{4^{l+1}}{(l+1)^2 {2l+2 \choose l+1}}$$

Simplifying, we obtain:

$$\sum_{m=0}^\infty (-1)^m \binom{l}{m} \frac{H_{m+1/2}+\log 4}{m+1} =\frac{2^{2l+1}}{(2l+1) (l+1) {2l \choose l}}$$

Multiplying by $x^l$ and summing from $0$ to $\infty$, we obtain:

$$\sum_{m=0}^\infty (-1)^m \frac{H_{m+1/2}+\log 4}{m+1} \frac{x^m}{(1-x)^{m+1}}= \sum_{l=0}^\infty \frac{2^{2l+1} x^l}{(2l+1) (l+1) {2l \choose l}}$$

After some simplifications, the right hand side gives us the Taylor series for $\frac{2}{x} \arcsin^2 \sqrt{x}$, or:

$$\frac{1}{x} \arcsin^2 \sqrt{x}=\sum_{l=0}^\infty \frac{(4 x)^l}{(2l+1) (l+1) {2l \choose l}}$$