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

Question

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?

Answer 1

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[
    {{nmax, TraditionalForm@Inactive[Sum][f[n], {n, 1, nmax}]}},
    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 *)

Answer 2

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

Answer 3

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