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How to approximate $$\sum_{n=1}^\infty\frac{{4n\choose 2n}\overline{H}_{2n}}{n 2^{4n}} ?$$ Where $\overline{H}_n=\sum_{k=1}^n \frac{(-1)^{k-1}}{k}$ is the skew harmonic number. The mathematica command for $\overline{H}_{2n}$ is $\log[2]-\text{LerchPhi}[-1,1,2n+1]$.

I tried Michael E2' command:

 major = Normal@Series[(Log[2] - LerchPhi[-1, 1, 2 n + 1]) Binomial[4 n, 
   2 n]/(n 2^(4 n)), {n, Infinity, 12}];
 majorsum = Sum[major, {n, Infinity}];
 majorsum + 
 NSum[(Log[2] - LerchPhi[-1, 1, 2 n + 1]) Binomial[4 n, 
  2 n]/(n 2^(4 n)) - major, {n, 1, Infinity}, NSumTerms -> 20, 
 WorkingPrecision -> 20, Method -> "WynnEpsilon"]

but it gave a result in terms of $n$ : enter image description here

which something unusual to see. Is there any other command or maybe we can do little changes to Michael E2's solution?

Thank you,

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  • $\begingroup$ How about a simple-minded approach N[Sum[Binomial[4 n, 2 n]*(Log[2] - LerchPhi[-1, 1, 2 n + 1])/n/ 2^(4 n), {n, 1, 2000}], 15] which results in 0.584900923610039? I leave an estimate of the rest on your own (or ask it in MSE). $\endgroup$ – user64494 Apr 6 at 6:04
  • $\begingroup$ N[Binomial[4 n, 2 n]*(Log[2] - LerchPhi[-1, 1, 2 n + 1])/n/2^(4 n) /. n -> 2000, 15] equals 3.09099743792782*10^-6. $\endgroup$ – user64494 Apr 6 at 6:14
  • $\begingroup$ Making use of that , N[Sum[Normal[ Series[Binomial[4 n, 2 n]*(Log[ 2] + (-1)^ n*(1/2)*(HarmonicNumber[-(1/2) + n/2] - HarmonicNumber[n/2]))/n/2^(4 n), {n, Infinity, 10}]], {n, 2001, Infinity}], 15] results in 0.0123649321640052. $\endgroup$ – user64494 Apr 6 at 6:19
  • $\begingroup$ It should be Log[2] - (-1)^n*LerchPhi[-1, 1, 1 + n] instead of Log[2] - LerchPhi[-1, 1, 1 + n] in your question. $\endgroup$ – user64494 Apr 6 at 12:35
  • $\begingroup$ @user64494 note that $(-1)^{2n}=1$. $\endgroup$ – Ali Shadhar Apr 7 at 2:21
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You can use a different representation of the skew harmonic numbers in terms of harmonic numbers: $$ \bar{H}_n=\frac{1}{2} (-1)^n \left(H_{\frac{n-1}{2}}-H_{\frac{n}{2}}\right)+\log (2) $$ This expression can be expanded around infinity

f[n_] := HarmonicNumber[(n - 1)/2] - HarmonicNumber[n/2]
Series[f[n], {n, Infinity, 6}]

$$-\frac{1}{n}+\frac{1}{2 n^2}-\frac{1}{4 n^4}+\frac{1}{2 n^6}+O\left({n}^{-7}\right)$$

and summed up

Sum[Binomial[4 n, 2 n]/(n 2^(4 n)) (Log[2] + (-1)^n 
    (-(1/(2 n)) + 1/(4 n^2) - 1/(8 n^4) + 1/(4 n^6))), 
    {n, Infinity}]
(*3/64 (4 HypergeometricPFQ[{1, 1, 1, 5/4, 7/4}, {3/2, 2, 2, 2}, -1]
-2 HypergeometricPFQ[{1, 1, 1, 1, 5/4, 7/4}, {3/2, 2, 2, 2, 2}, -1] 
+ HypergeometricPFQ[{1, 1, 1, 1, 1, 1, 5/4, 7/4}, {3/2, 2, 2, 2, 2, 2, 2}, -1]
-2 HypergeometricPFQ[{1, 1, 1, 1, 1, 1, 1, 1, 5/4, 7/4}, {3/2, 2, 2, 2, 2, 2, 2, 2, 2}, -1] 
+8 HypergeometricPFQ[{1, 1, 5/4, 7/4}, {3/2, 2, 2}, 1] Log[2])*)
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  • $\begingroup$ The result of Series is not a good approximation for f[n] if n is small. Your calculation is not correct:N[3/64 (4 HypergeometricPFQ[{1, 1, 1, 5/4, 7/4}, {3/2, 2, 2, 2}, -1] - 2 HypergeometricPFQ[{1, 1, 1, 1, 5/4, 7/4}, {3/2, 2, 2, 2, 2}, -1] + HypergeometricPFQ[{1, 1, 1, 1, 1, 1, 5/4, 7/4}, {3/2, 2, 2, 2, 2, 2, 2}, -1] - 2 HypergeometricPFQ[{1, 1, 1, 1, 1, 1, 1, 1, 5/4, 7/4}, {3/2, 2, 2, 2, 2, 2, 2, 2, 2}, -1] + 8 HypergeometricPFQ[{1, 1, 5/4, 7/4}, {3/2, 2, 2}, 1] Log[2])] is 0.72777.See my comments to the question. $\endgroup$ – user64494 Apr 6 at 10:43
  • $\begingroup$ I did my answer, having collected my comments. Don't hesitate to ask for further explanation in need. $\endgroup$ – user64494 Apr 6 at 10:56
  • $\begingroup$ I wonder the upvoter: the answer is wrong as explained in my comments. $\endgroup$ – user64494 Apr 6 at 11:51
  • $\begingroup$ @user64494 I voted for the different representation of the skew harmonic number. $\endgroup$ – Chip Hurst Apr 6 at 12:06
  • $\begingroup$ @ChipHurst: That was noticed in my comment to the question before the yarchik's answer. $\endgroup$ – user64494 Apr 6 at 12:19
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We can use Michael E2's method with a different representation for the skew Harmonic number:

f[n_] = (Log[2] + (-1)^n/2 (HarmonicNumber[(n - 1)/2] - HarmonicNumber[n/2])) Binomial[4n, 2n]/(n*2^(4n));

major = Normal@Series[f[n], {n, Infinity, 12}];
majorsum = Sum[major, {n, Infinity}];
approxtail = N[Sum[f[n] - major, {n, 1, 500}], 30];

majorsum + approxtail
0.79573336271051611600770246918
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First, we directly calculate the sum of the first 2000 terms:

N[Sum[Binomial[4 n, 2 n]*(Log[2] - LerchPhi[-1, 1, 2 n + 1])/n/     2^(4 n), {n, 1, 2000}], 15]

0.584900923610039

Second, making use of that answer by Andreas,

N[Sum[Normal[Series[Binomial[4 n,2 n]*(Log[2] + (-1)^n*(1/2)*(HarmonicNumber[-(1/2) + n/2] - 
HarmonicNumber[n/2]))/n/2^(4 n), {n, Infinity, 10}]], {n,2001, Infinity}], 15]

0.0123649321640052

Finally, the sum under consideration approximately equals 0.0123649321640052 + 0.584900923610039.

Addition. There is a substantial difference between

N[Sum[Binomial[4 n, 2 n]*(Log[2] - LerchPhi[-1, 1, 2 n + 1])/n/
2^(4 n), {n, 1, 2000}], 15]

0.584900923610039

and

N[Sum[Binomial[4 n, 
2 n]*(Log[
    2] + (-1)^
     n*(1/2)*(HarmonicNumber[-(1/2) + n/2] - 
      HarmonicNumber[n/2]))/n/2^(4 n), {n, 1, 2000}], 15]

0.783368430546511

The OP writes an incorrect formula Log[2] - LerchPhi[-1, 1, 1 + n] instead of the correct one Log[2] - (-1)^n*LerchPhi[-1, 1, 1 + n] in the question.

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