Finding $\displaystyle \sum \limits _{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n(2)^n \binom{2n}{n}}$

I want to find the closed form of:

$$\displaystyle \tag*{} \sum \limits _{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n(2)^n \binom{2n}{n}}$$

Where $$H_{k}$$ is $$k^{\text{th}}$$ harmonic number

Can anyone tell me the value of the sum using Mathematica?

• Welcome to the Mathematica Stack Exchange. I think that your question can be more suitably addressed over at the Math Stack Exchange. This stack site is about the technical software called Mathematica and the associated Wolfram Language. Best of luck.
– Syed
Apr 7, 2022 at 13:39
• @Syed Thank you, I am aware of that. I don't have access to Mathematica, so can you input my sum into the software and tell me the value, please? Apr 7, 2022 at 13:52
• @MarcoB Thank you. If you have access to Mathematica, can you find the value of my sum, please? Apr 7, 2022 at 13:52

Clear["Global*"]

f[n_] := (-1)^
n (HarmonicNumber[2 n] - HarmonicNumber[n])/(n*2^n*Binomial[2 n, n])


The terms of the sum converge to zero rapidly

DiscretePlot[Abs[f[n]], {n, 1, 25},
ScalingFunctions -> "Log"]


The partial sums are

(tab = Join[
Table[{nmax, Sum[f[n], {n, 1, nmax}]}, {nmax, 1, 28}] /.
r_Rational :> N[r, 25]]) //
Grid[#, Frame -> All] &


The partial sum expressed as a DifferenceRoot is

psum[nmax_] = Sum[f[n], {n, 1, nmax}]


Using AskConstants to identify a constant with this approximate numeric value

(sum = (Log[2]^2 - Pi^2/12)/3) // TraditionalForm


Checking,

Block[{$MaxExtraPrecision = 500}, Join[ {{nmax, TraditionalForm[sum - Inactive[Sum][f[n], {n, 1, nmax}]]}}, Table[{nmax, N[sum - Sum[f[n], {n, 1, nmax}], nmax + 20] // N}, {nmax, 20, 200, 20}]] // Grid[#, Frame -> All] &]  EDIT: An alternate approach (sum2 = FullSimplify@ Total[Sum[#, {n, 1, Infinity}] & /@ Assuming[n ∈ PositiveIntegers, (List @@ (f[n] // FunctionExpand // Simplify // Expand))]]) // TraditionalForm  N[sum2, 20] (* -0.11400467316863726452 *)  Checking that sum and sum2 are equivalent Block[{$MaxExtraPrecision = 500},
N[sum - sum2, 200]]

(* N::meprec: Internal precision limit \$MaxExtraPrecision = 500. reached while evaluating 1/3 (-(π^2/12)+Log[2]^2)+1/72 (7 π^2-3 (9 Log[<<1>>]^2+8 PolyLog[2,Power[<<2>>]]-8 PolyLog[2,Times[<<2>>]])). *)

(* 0.*10^-699 *)

• Wow, I am sorry I couldn't do that green tick as I already did. But your answer almost helped me solve the questions mathematically :). Apr 7, 2022 at 15:53
• @4444 to be fair, you can accept Bob's answer and unaccept mine. Just click on the green checkmark. It's more thorough compared to mine. Oh, and Bob, I did not know about the AskConstants so many thanks for demonstrating its use
– bmf
Apr 7, 2022 at 15:54
• @bmf thanks for your permission. And I am very grateful for the 'AskConstants' Apr 7, 2022 at 15:57
• @4444 it's not a matter of permission. Bob is an excellent user with high-quality answers and this answer here might also be helpful for future users :)
– bmf
Apr 7, 2022 at 15:57
• Wolfram alpha also attempts to find closed forms from numerical numbers but it did not find that one. From this one example, it seems to me that the algorithm used is too focused on searching for Formulas with algebraic numbers, Pi and Exp[1]. Aug 5, 2022 at 1:04
f[n_] := (((-1)^n (HarmonicNumber[2 n] - HarmonicNumber[n]))/(
n 2^n Binomial[2 n, n]))


First, we check the sum convergence

SumConvergence[f[n], n]


Doing the sum analytically:

Sum[f[n], {n, 1, Infinity}]


that takes too long for the time I can spare. So, let's do a couple of values and take it from there.

Table[Sum[f[n], {n, 1, xx}], {xx, 1, 21}]


You can try to find an analytic formula for the numerator and denominator of the above data, either by using FindSequenceFunction or OEIS. Neither leads to an answer.

So, finally we resort to numerics

Table[Sum[f[n], {n, 1, xx}], {xx, 1, 21}] // N


and we see that it quickly converges to -0.114005

Let $$a_n=\frac{ H_{2n}-H_{n}}{n\,2^n\, \binom{2n}{n}}\implies \frac{a_{n+1}}{a_n}=\frac{n \left(H_{n+1}-H_{2 n+2}\right)}{4 (2 n+1) \left(H_n-H_{2 n}\right)}$$ Using asymptotics $$\frac{a_{n+1}}{a_n}=\frac{1}{8}-\frac{1}{16 n}+\frac{1+\log (2)}{32 n^2 \log (2)}+O\left(\frac{1}{n^3}\right)$$ So, we can expect a quite fast convergence.

Facing an alternating series,

$$\sum_{n=1}^\infty (-1)^n a_n=\sum_{n=1}^p (-1)^n a_n+\sum_{n=p+1}^\infty (-1)^n a_n$$ consider $$a_{p+1}$$ $$H_{2(p+1)}-H_{p+1}=\log (2)-\frac{1}{4 n}+\frac{5}{16 n^2}-\frac{3}{8 n^3}+O\left(\frac{1}{n^4}\right)$$

$$\log\big[H_{2(p+1)}-H_{p+1}\big]=\log (\log (2))-\frac{1}{4p \log (2)}+\frac{10 \log (2)-1}{32 p^2 \log ^2(2)}+O\left(\frac{1}{p^3}\right)$$ $$\log\Bigg[2^{p+1} (p+1) \binom{2 p+2}{p+1}\Bigg]=3p \log (2)+\frac{1}{2} \log \left(\frac{64 p}{\pi }\right)+\frac{3}{8 p}-\frac{1}{8 p^2}+O\left(\frac{1}{p^3}\right)$$

$$\log(a_{p+1})=-3p \log (2)+\left(\log (\log (2))-\frac{1}{2} \log \left(\frac{64 p}{\pi }\right)\right)+O\left(\frac{1}{p}\right)$$ So, if we want $$a_{p+1} \leq \epsilon \quad \implies \quad p \geq \frac{W(t)}{6 \log (2)} \qquad \text{where} \qquad t=\frac{3 \pi \log ^3(2)}{32 \epsilon ^2}$$ $$W(t)$$ being Lambert function.

Suppose $$\epsilon=10^{-16}$$, this gives as a real $$p=16.1471$$ (notice that the exact solution is $$16.1263$$.

Computing for $$p=17$$ gives for the summation $$-\frac{6773770929644673431621962072907}{59416607594890592064865566720000}$$ which, in decimal, is $$-0.11400467316863727867$$.

Now,being lazy, I went here; click the last button ("Integer Relation Algorithms") and you will see the result already given

• It's nice to see a mathematical proof as well!
– hana
Apr 8, 2022 at 16:22

Edit

After writing the answer below I found from suggestions that an analytical answer was already found here.

Reformulating the sum as an integral

A well-known method to compute integrals is to replace the integrand by a series and to integrate term by term. We will do the opposite here. We will replace the terms in the series with an integral.

The Wikipedia page on harmonic numbers gives an integral representation which we will use:

$$H_n=\int_{0}^{1} \frac{1-t^{n}}{1-t} d t$$

Hence, the series above can be written as:

$$\sum \limits _{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n(2)^n \binom{2n}{n}}=\int_{0}^{1}\frac{1}{1-t}\sum \limits _{n=1}^{\infty} \frac{(-1)^n(t^n-t^{2n})}{n2^n \binom{2n}{n}} d t$$

An intermediate problem is then how to compute:

$$f(t)=\sum \limits _{n=1}^{\infty} \frac{t^n}{n\binom{2n}{n}}$$

Mathematica shows that this function is :

$$f(t)=\frac{2 \sqrt{t} \operatorname{ArcSin}\left[\frac{\sqrt{t}}{2}\right]}{\sqrt{4-t}}$$

Substituting $$t$$ for $$t^2$$ one could probably verify this by showing that both expressions verify the same linear differential equation of order 2 (I used Mathematica's DifferentialRootReduce to find the differential equation from the explicit ArcSin form), however, that sounds a bit tedious. Instead, I will derive a different representation for $$f$$ using the following integral representation given on the Wikipedia page for the beta function :

$$\frac{1}{\binom{2n}{n}} =\frac{n}{2} \operatorname{Beta}(n,n) = \frac{n}{2} \int_0^1 (1-x)^{n-1} x^{n-1} \mathrm{d} x$$

The function f above is then represented by :

$$f(t)=\int_0^1\frac{t}{2} \sum \limits _{n=1}^{\infty} \left(tx(1-x)\right)^{n-1} \mathrm{d} x =\int_0^1 \frac{t}{2(1-(xt(1-x)))} \mathrm{d} x$$

This integral can be computed by obtaining the canonical form of the second-degree polynomial in the denominator and then using a change of variables to obtain an ArcTan function. I suppose that using trigonometry there is a way to go from the ArcTan version to the ArcSin version of Mathematica.

The original sum involving harmonic numbers can be written as:

$$\sum \limits _{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n(2)^n \binom{2n}{n}}=\int_{0}^{1}\frac{1}{1-t}\left(f(-t/2)-f(-t^2/2)\right) \mathrm{d} t$$

The singularity at $$t=1$$ in the denominator is canceled by $$f(-1/2)$$-$$f(-1/2)$$ in the numerator.

Using the definition of $$f$$ in terms of the ArcSin function, Mathematica seems to be able to compute such integrals (Mathematica seems to have trouble when the boundaries are specified so I took an indefinite integral and computed limits using Series).

Edit:

I have been trying to compute the integral on my own but the computation seems quite tedious.

Some steps to put the integral into a more simple/canonical form :

• Rewrite $$f(-t/2)$$ and $$f(-t^2/2)$$ in terms of Arcsinh instead of ArcSin

• Compute the two terms containing $$f(-t/2)$$ and $$f(-t^2/2)$$ separately

• Change of variables $$t=4b^2$$ to remove the Sqrt in ArcSinh in $$f(-t/2)$$

• Notice that after the substitution $$t=b^2$$, $$f$$ looks like $$\frac{d}{db}\operatorname{ArcSin}(b)^2$$ and use that in an integration by parts.

• Remove the ArcSinh by using the change of variables $$b=\operatorname{Sinh}(a)$$

• Write Cosh and Sinh in terms of Exp and do a change of variables $$\operatorname{Exp}(a)=c$$

• The integral will look like a rational function time the square of Logarithm for both the $$f(-t/2)$$ and $$f(-t^2/2)$$ if I remember well (I might be wrong)

• Try to use integral representations of the dilograrithm to obtain the final result. Integration by parts might be needed where one of the logarithms in $$\log(g)^2$$ is differentiated. I did not try this last step and maybe one of the integral representations of the dilogarithm is close to the integral from the step before. I do not remember if other polylogarithms are needed.

Resources for integral representations

• Wikipedia

• Wolfram Alpha (write the function in the search box, you might have to click on the link above the result from the search where it is written "Use as a math function instead" ). You might have to click on more to see more representations of the function.

• NIST Digital Library of Mathematical Functions