# 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 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 at 13:52
• @MarcoB Thank you. If you have access to Mathematica, can you find the value of my sum, please? Apr 7 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 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 at 15:54
• @bmf thanks for your permission. And I am very grateful for the 'AskConstants' Apr 7 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 at 15:57
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 at 16:22