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Help finding this sum.

$$\sum_{n=0}^{\infty}(-1)^{n}\left(\frac{1}{2}-c(2n+1)\Phi(-1,1,c\left(2n+1\right)+1)\right)$$

Where $\Phi$ is the Lerch Transcendent. The sum can then be written as

$$\sum_{n=0}^{\infty}\left(-1\right)^{n}\left(\frac{1}{2}-c\left(-1\right)^{c}\left(2n+1\right)\left(\ln2-\sum_{k=0}^{c\left(2n+1\right)-1}\frac{\left(-1\right)^{k}}{k+1}\right)\right)$$

Partial sums include a fraction term and a $\ln2$ term. And the limit numerically appears to converge. Can we find a closed form?

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    $\begingroup$ I am skeptical about a possible closed form. Does my upper bound match your numerical results ? $\endgroup$ Feb 17, 2020 at 16:41
  • $\begingroup$ Yes it numerically matches, though not an upper bound. For example, I tried $c=20$. $\endgroup$
    – tyobrien
    Feb 17, 2020 at 16:45
  • $\begingroup$ For $c=20$, the result is $0.00980241$; for $c=200$, it is $0.000981733$;. for $c=2000$, it is $0.0000981746$. $\frac {41}{208\,c}$ is probaly much better. For $c=20$, it would give $0.00985577$. Do you want a better constant ? $\endgroup$ Feb 17, 2020 at 16:54
  • $\begingroup$ Oh, I made an error. Looks like a fair upper bound. $\endgroup$
    – tyobrien
    Feb 17, 2020 at 16:55
  • $\begingroup$ Neat, but that won't be necessary ;) $\endgroup$
    – tyobrien
    Feb 17, 2020 at 17:01

1 Answer 1

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This is not an answer.

What it seems to me is that an upper bound of $$S(c)=\sum_{n=0}^{\infty}(-1)^{n}\left(\frac{1}{2}-c(2n+1)\Phi(-1,1,c\left(2n+1\right)+1)\right) \sim \frac{1}{5 \,c}$$

A good approximation seems to be $$S(c)=\frac{269}{1370 c}-\frac{29}{244 c^3}$$

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