An interesting sum involving binomial coefficients $\sum_{k=1}^n\frac{(-1)^{k+1} 2^{2k}}{k}\binom{n}{k}\binom{2k}{k}^{-1}$ 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:

*

*The binomial transform at the level of (exponential) generating series is very powerful. I have to get more comfortable with formulas.


*There are other interesting formulas that use the Beta integrals to express the inverse of binomial coefficients as an integral--- very useful.


*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 $m$.
 A: 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:
$$\frac{-2\sqrt{-x}\arctan\sqrt{-x}}{1-x}=\sum_{n=1}^\infty x^n\sum_{k=1}^n\frac2{2k-1}$$
$$=\sum_{n=1}^\infty x^n\left(\sum_{k=1}^{2n}\frac2k-\sum_{k=1}^n\frac1k\right)=\sum_{n=1}^\infty(\color{blue}{2H_{2n}-H_n})x^n$$
$$\color{blue}{\sum_{k=1}^n\frac{(-1)^{k+1}2^{2k}}{k\binom{2k}k}\binom nk=2H_{2n}-H_n}$$
A: $$\sum_{k=1}^n(-1)^{k-1}\binom{n}{k}\frac{4^k}{k \binom{2k}{k}}=\sum_{k=1}^n(-1)^{k-1}\binom{n}{k}\left(2\int_0^{\pi/2}\sin^{2k-1}(x)\mathrm{d}x\right)$$
$$=-2\int_0^{\pi/2}\frac{1}{\sin x}\left(\sum_{k=1}^n \binom{n}{k}(-\sin^2 x)^k\right)\mathrm{d}x$$
$$=-2\int_0^{\pi/2}\frac{1}{\sin x}\left(-1+\cos^{2n}x\right)\mathrm{d}x$$
$$=2\int_0^1\frac{1}{1-t^2}(1-t^{2n})\mathrm{d}t$$
$$=2\int_0^1\left(\frac{1}{1-t}-\frac{t}{1-t^2}\right)(1-t^{2n})\mathrm{d}t$$
$$=2\int_0^1\frac{1-t^{2n}}{1-t}\mathrm{d}t-\int_0^1\frac{2t(1-t^{2n})}{1-t^2}\mathrm{d}t$$
$$=2\int_0^1\frac{1-t^{2n}}{1-t}\mathrm{d}t-\int_0^1\frac{1-u^n}{1-u}\mathrm{d}u$$
$$=2H_{2n}-H_n$$
A: Let $$\sum_{k=1}^n\frac{(-1)^{k+1} 2^{2k}}{k}\binom{n}{k}\binom{2k}{k}^{-1},$$
Use $\frac{1}{N \choose K}=(N+1)\int_{0}^{1} x^K (1-x)^{N-K} dx$
Then $$S=\sum_{k=1}^n\int_0^1(2k+1)\frac{(-1)^{k+1}2^{2k}}k\binom nk x^k(1-x)^k\,dx$$
$$=-\int_0^1\sum_{k=1}^n\frac{2k+1}k\binom nk(-4x(1-x))^k\,dx$$
$$=-2\int_0^1\sum_{k=1}^n\binom nk(4x^2-4x)^k\,dx-\int_0^1\sum_{k=1}^n\binom nk\frac{(4x^2-4x)^k}k\,dx$$
$$=-2\int_0^1((2x-1)^{2n}-1)\,dx-\int_{0}^{1}\sum_{k=1}^n\binom nk\frac{(4x^2-4x)^k}k\,dx$$
$$=\frac{4n}{2n+1}-\int_0^1\sum_{k=1}^n\binom nk\frac{((2x-1)^2-1)^k}k\,dx$$
Note that $\int \sum_{k=1}^{n} {n \choose k} (-z)^{k-1}dz= \int\frac{(1-z)^n-1}{(1-z)-1}=-\sum_{k=1}^{n} {n \choose k}\frac{(-z)^k}{k}.$ Let $(1-2x)=t$, then
$$S=\frac{4n}{2n+1}-\int_{0}^{
1} \frac{(1-2x)^{2n}-1}{(1-2x)^2-1}dx=\frac{4n}{2n+1}+2\int_{0}^{1} \frac{t^{2n}-1}{t^2-1}dt=\frac{4n}{2n+1}+2 \sum_{k=1}^n \frac{1}{2k+1}$$
$$\implies S= 2\sum_{k=0}^{n-1} \frac{1}{2k+1}=2H_{2n}-H_n.$$
See $$1+1/3+1/5+\cdots+1/(2n-1)=(1+1/2+1/3+\cdots+1/(2n))-(1/2+1/4+1/6+\cdots+1/(2n))=H_{2n}-\frac{1}{2}H_n.$$
A: We seek to verify that
$$\sum_{k=1}^n \frac{(-1)^{k+1} 2^{2k}}{k}
{n\choose k} {2k\choose k}^{-1}
= 2 H_{2n} - H_n.$$
Recall from MSE
4316307
the following identity which was proved there: with $1\le k\le n$
$$\frac{1}{k} {n\choose k}^{-1}
= [z^n] \log\frac{1}{1-z} (z-1)^{n-k}.$$
We get for our sum
$$\sum_{k=1}^n (-1)^{k+1} 2^{2k} {n\choose k}
[z^{2k}] \log\frac{1}{1-z} (z-1)^k
\\ = (-1)^n 2^{2n}
\sum_{k=0}^{n-1} {n\choose k} (-1)^{k+1} 2^{-2k}
[z^{2n-2k}] \log\frac{1}{1-z} (z-1)^{n-k}
\\ = (-1)^{n+1} 2^{2n} [z^{2n}] \log\frac{1}{1-z} (z-1)^n
\sum_{k=0}^{n-1} {n\choose k} (-1)^{k} 2^{-2k}
z^{2k} (z-1)^{-k}.$$
We see that we may raise $k$ to $n$ because this is a zero
contribution owing to the fact that $\log\frac{1}{1-z}$ does not have a constant term, getting
$$(-1)^{n+1} 2^{2n} [z^{2n}] \log\frac{1}{1-z} (z-1)^n
\left[1-\frac{z^2}{4(z-1)}\right]^n
\\ = (-1)^{n+1} [z^{2n}] \log\frac{1}{1-z}
\left[4z-4-z^2\right]^n
\\ = - [z^{2n}] \log\frac{1}{1-z} (z-2)^{2n}.$$
This is
$$- \; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{2n+1}} \log\frac{1}{1-z} (z-2)^{2n}.$$
Now put $z/(z-2) = v$ so that $z=2v/(v-1)$ and $dz = - 2/(v-1)^2 \; dv$
to obtain
$$\; \underset{v}{\mathrm{res}} \;
\frac{1}{v^{2n+1}} \log\frac{1}{1-2v/(v-1)} \frac{v-1}{2}
2 \frac{1}{(v-1)^2}
\\ = - \; \underset{v}{\mathrm{res}} \;
\frac{1}{v^{2n+1}} \frac{1}{1-v}
\log\frac{v-1}{-v-1}
\\ = - \; \underset{v}{\mathrm{res}} \;
\frac{1}{v^{2n+1}} \frac{1}{1-v}
\log\frac{1-v}{1+v}.$$
We get two pieces, the first is
$$\; \underset{v}{\mathrm{res}} \;
\frac{1}{v^{2n+1}} \frac{1}{1-v}
\log\frac{1}{1-v} = H_{2n}.$$
The second is
$$- \; \underset{v}{\mathrm{res}} \;
\frac{1}{v^{2n+1}} \frac{1}{1-v}
\log\frac{1}{1+v}
= - \sum_{q=1}^{2n} \frac{(-1)^q}{q}
= - \left[\sum_{p=1}^n \frac{1}{2p}
- \sum_{p=0}^{n-1} \frac{1}{2p+1} \right]
\\ = - \frac{1}{2} H_n
+ H_{2n} - \sum_{q=1}^n \frac{1}{2q}
= H_{2n} - H_n.$$
Collecting the two pieces we obtain
$$\bbox[5px,border:2px solid #00A000]{
2 H_{2n} - H_n}$$
as claimed.
