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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:

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

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

  3. 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 $m$.

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    $\begingroup$ It is remarkable that Mathematica calculates the sum almost immediately with the result $s=2 H(2n)-H(n)$. Correspondingly, there should be a very simple way to derive it. But I haven't found it yet. $\endgroup$ Commented Oct 8, 2022 at 22:49
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    $\begingroup$ What it really gives in the first place is the elegant expression $\psi ^{(0)}\left(n+\frac{1}{2}\right)-\psi ^{(0)}\left(\frac{1}{2}\right) (=H_{n-\frac{1}{2}}-H_{-\frac{1}{2}})$. The relation is of the general type $\sum _{k=1}^n \frac{1}{a+k-1}=\psi ^{(0)}(a+n)-\psi ^{(0)}(a)$ $\endgroup$ Commented Oct 9, 2022 at 5:56
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    $\begingroup$ Thank you for the TOC. Another simplification: the most simple form of our sum I could find is: $s_n=H_{n-\frac{1}{2}}+\log (4)$. Now there's almost nothing left ;-) $\endgroup$ Commented Oct 10, 2022 at 14:41
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    $\begingroup$ @Dr.WolfgangHintze Given that Oleg Marychev had worked on Mathematica for a long time, maybe not THAT remarkable. $\endgroup$
    – Igor Rivin
    Commented Oct 11, 2022 at 2:05
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    $\begingroup$ @ Igor Rivin I am grateful to all developers of Mathematica even if there are sums which take remakably long to be evaluated. $\endgroup$ Commented Oct 12, 2022 at 4:26

4 Answers 4

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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: $$\frac{-2\sqrt{-x}\arctan\sqrt{-x}}{1-x}=\sum_{n=1}^\infty x^n\sum_{k=1}^n\frac2{2k-1}$$ $$=\sum_{n=1}^\infty x^n\left(\sum_{k=1}^{2n}\frac2k-\sum_{k=1}^n\frac1k\right)=\sum_{n=1}^\infty(\color{blue}{2H_{2n}-H_n})x^n$$ $$\color{blue}{\sum_{k=1}^n\frac{(-1)^{k+1}2^{2k}}{k\binom{2k}k}\binom nk=2H_{2n}-H_n}$$

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  • $\begingroup$ Very neat! Thank you! $\endgroup$
    – orangeskid
    Commented Oct 7, 2022 at 6:33
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$$\sum_{k=1}^n(-1)^{k-1}\binom{n}{k}\frac{4^k}{k \binom{2k}{k}}=\sum_{k=1}^n(-1)^{k-1}\binom{n}{k}\left(2\int_0^{\pi/2}\sin^{2k-1}(x)\mathrm{d}x\right)$$

$$=-2\int_0^{\pi/2}\frac{1}{\sin x}\left(\sum_{k=1}^n \binom{n}{k}(-\sin^2 x)^k\right)\mathrm{d}x$$

$$=-2\int_0^{\pi/2}\frac{1}{\sin x}\left(-1+\cos^{2n}x\right)\mathrm{d}x$$

$$=2\int_0^1\frac{1}{1-t^2}(1-t^{2n})\mathrm{d}t$$

$$=2\int_0^1\left(\frac{1}{1-t}-\frac{t}{1-t^2}\right)(1-t^{2n})\mathrm{d}t$$

$$=2\int_0^1\frac{1-t^{2n}}{1-t}\mathrm{d}t-\int_0^1\frac{2t(1-t^{2n})}{1-t^2}\mathrm{d}t$$

$$=2\int_0^1\frac{1-t^{2n}}{1-t}\mathrm{d}t-\int_0^1\frac{1-u^n}{1-u}\mathrm{d}u$$

$$=2H_{2n}-H_n$$

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  • $\begingroup$ Very nice! Second row I am sure it's $(-\sin^2 x)^k$. So the idea is to express the factor without $\binom{n}{k}$ as a moment. Great! $\endgroup$
    – orangeskid
    Commented Jan 30, 2023 at 1:41
  • $\begingroup$ Oh thank you ... yes I'll fix it $\endgroup$ Commented Jan 30, 2023 at 1:44
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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.

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  • $\begingroup$ Amazing! Need to chew on it... $\endgroup$
    – orangeskid
    Commented Oct 7, 2022 at 21:02
  • $\begingroup$ You directed in your answer to the other very interesting question, it is all very useful, thank you! $\endgroup$
    – orangeskid
    Commented Oct 11, 2022 at 1:41
  • $\begingroup$ You're welcome. $\endgroup$ Commented Oct 11, 2022 at 19:37
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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 $$1+1/3+1/5+\cdots+1/(2n-1)=(1+1/2+1/3+\cdots+1/(2n))-(1/2+1/4+1/6+\cdots+1/(2n))=H_{2n}-\frac{1}{2}H_n.$$

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  • $\begingroup$ Very nice and informative! Thank you! $\endgroup$
    – orangeskid
    Commented Oct 7, 2022 at 16:32

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