The following sum appears in Brychkov, Marichev, and Prudnikov, Integrals and Series, Vol 1, 4.2.8, #25
$$\sum_{k=1}^n\frac{(-1)^{k+1} 2^{2k}}{k}\binom{n}{k}\binom{2k}{k}^{-1}= 2 H_{2n} – H_n$$
where $H_n$ are the harmonic numbers.
Here is what I've tried:
We have the series expansion:
$$\sum_{n\ge 1} \frac{2^{2n-1} }{n \binom{2n}{n}} x^{2n-1} = \frac{\arcsin x}{\sqrt{1-x^2}}$$
and so
$$\sum_{n\ge 1} \frac{2^{2n-1} }{n \binom{2n}{n}} x^{n} = \frac{\sqrt{x} \arcsin \sqrt{x}}{\sqrt{1-x}}$$
Now we can use the relation between the generating functions for the binomial transform
$$G(x) = \frac{1}{1-x}\cdot F(-\frac{x}{1-x})$$
Now things get a bit fuzzy : it's not clear what will be the expansion after the transformation, and how that involves harmonic numbers.
Also, here is the Taylor expansions involving harmonic numbers
$$\sum_{n\ge 1} H_n x^n= \frac{1}{1-x} \log \frac{1}{1-x}$$
and from here using $x \mapsto \pm \sqrt{x}$ and averaging, we can get the series
$$\sum_{n \ge 1} H_{2n} x^n $$
Any feedback is appreciated!
$\bf{Added:}$ Thank you all for all the great answers! I've learned a lot.
I will try to share a part of what I've learned from your answers:
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The binomial transform at the level of (exponential) generating series is very powerful. I have to get more comfortable with formulas.
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There are other interesting formulas that use the Beta integrals to express the inverse of binomial coefficients as an integral— very useful.
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I've learned a new formula from this question, indicated by @Marko Riedel:
We have the identity in $\alpha$
$$\sum_{k=1}^n \frac{\binom{\alpha + n-k}{n-k}}{\binom{\alpha + n}{n}} \cdot \frac{1}{k} = \sum_{k=1}^n \frac{1}{\alpha + k}$$
This can also be rewritten as
$$\sum_{k=1}^n \frac{\binom{n}{k}}{\binom{\alpha + n}{k}} \cdot \frac{1}{k} = \sum_{k=1}^{n} \frac{1}{\alpha + k}$$
or with $\alpha = \beta- n$,
$$\sum_{k=1}^n \binom{n}{k} \frac{1}{k \binom{\beta}{k} } = \sum_{k=0}^{n-1} \frac{1}{\beta- k}$$
If we take $\alpha = -\frac{1}{2}-n$ in the formula we get our formula ( we have
$\binom{-\frac{1}{2}}{k} = \frac{(-1)^k \binom{2k}{k}}{2^{2k}}$)
$\bf{Added:}$ This formula (slightly modified) at 4.2.8. #27 appears in the Volume but only for natural values of $m$.
Best Answer
Perform the binomial transform (noting that the zeroth term of the original sequence is $0$) and simplify: $$\sum_{n=1}^\infty\frac{2^{2n-1}}{n\binom{2n}n}x^n=\frac{\sqrt x\arcsin\sqrt x}{\sqrt{1-x}}$$ $$\sum_{n=1}^\infty x^n\sum_{k=1}^n\frac{(-1)^k2^{2k-1}}{k\binom{2k}k}\binom nk=\frac1{1-x}\frac{\sqrt{-x/(1-x)}\arcsin\sqrt{-x/(1-x)}}{\sqrt{1+x/(1-x)}}$$ $$=\frac{\sqrt{-x}\arctan\sqrt{-x}}{1-x}$$ $$\sum_{n=1}^\infty x^n\color{blue}{\sum_{k=1}^n\frac{(-1)^{k+1}2^{2k}}{k\binom{2k}k}\binom nk}=\frac{-2\sqrt{-x}\arctan\sqrt{-x}}{1-x}$$ The series expansion of the numerator here is $$-2\sqrt{-x}\arctan\sqrt{-x}=\sum_{n=1}^\infty\frac2{2n-1}x^n$$ so the Cauchy product eventually leads to the desired harmonic sum: $$\frac{-2\sqrt{-x}\arctan\sqrt{-x}}{1-x}=\sum_{n=1}^\infty x^n\sum_{k=1}^n\frac2{2k-1}$$ $$=\sum_{n=1}^\infty x^n\left(\sum_{k=1}^{2n}\frac2k-\sum_{k=1}^n\frac1k\right)=\sum_{n=1}^\infty(\color{blue}{2H_{2n}-H_n})x^n$$ $$\color{blue}{\sum_{k=1}^n\frac{(-1)^{k+1}2^{2k}}{k\binom{2k}k}\binom nk=2H_{2n}-H_n}$$