The right numerical value of the closed form of $\sum_{n=1}^\infty\frac{4^n H_n^2}{{2n\choose n}n^2}$ is $40.66752074791188333...$.
I tried to verify this result on Mathematica using the command:
NSum[4^n HarmonicNumber[n]^2/(Binomial[2 n, n] n^2), {n, 1, Infinity},
WorkingPrecision -> 8]
but it failed to give any approximation, so I replaced Infinity
by 1000 000
and it gave 39.70 which is close to the numeric value of the closed form. My question is: why Mathemtica fails to give the right approximation for series involving the binomial series when we use Infinity
?. Other question is: is there another command besides
NSum[4^n HarmonicNumber[n]^2/(Binomial[2 n, n] n^2), {n, 1, 1000000},
WorkingPrecision -> 8]
that gives a more accurate approximation?
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