This question is related to my question here which it were related to irrationality inverse square root of Gamma function , I plug the following sum : $$\sum _{n=1}^{\infty } \frac{1}{\sqrt{\Gamma \left(n^5\right)}}$$in Wolfram Alpha with precision $30$ ,it returned the sum to exactly $1$, I have used The standard definition of $\Gamma(n) =(n-1)!$ trying to get its partial sum but it were complicated to me for evaluation, Now ,Is that sum interpreted any standard result ? and how I can evaluate it since it is rational ?
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8The first term is 1, and the other terms are absolutely miniscule and don't show up with 30 decimal places, but they are non-zero. – Ragib Zaman Jan 04 '20 at 22:18
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3PARI/GP computes $1.0000000000000000110278059538310609033$ for the sum of the first 10 terms. – Martin R Jan 04 '20 at 22:19
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1bc -l gets .0000000000000000110278059538310609032942060847778971258447535062193870006468053968418970017788817337 as the 2nd term. – kimchi lover Jan 04 '20 at 22:44
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4Asking many times the same question doesn't change the answer – reuns Jan 04 '20 at 23:24
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@kimchilover. I agree with you. Suming the first, second and third terms gives $1000$ exact decimal figures. – Claude Leibovici Jan 05 '20 at 05:08
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@kimchilover Nice to see someone using bc:) – Klangen Jan 07 '20 at 07:23
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As mentioned in the comment section,
$$\sum _{n=1}^{\infty } \frac{1}{\sqrt{\Gamma \left(n^5\right)}} \neq 1.$$
Adding the first $10$ terms together, we get
$$1.0000000000000000110278059538310609032942060847778971258447535062193870006468053968418970017788817337127$$
You can use Alpha to convince yourself.
Klangen
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