Does this sum have a closed form?

I was working with the identity $\displaystyle\frac{e+e^\zeta +e^{\zeta ^2}+...+e^{\zeta ^{n-1}}}{n}-1=\sum_{k=1}^{\infty}\frac{1}{(nk)!}$ , which can be proven by using the Taylor series for e^x and plugging in the nth roots of unity (zeta is a primitive nth root here).By summing the identity for all n, it in fact becomes $\displaystyle\sum_{n=1}^{\infty}\frac{1}{\tau(n).n!}$ , which is a much more natural sum to consider (here tau is the divisor function).Since $1\leq\tau(n)<n+1$ , we can show that the sum is between e-2 and e-1.

I was wondering if we could find an asymptotic for the partial sums or indeed find a formula for the infinite sum of any of these (the sum of the powers of e or the tau one).

If anyone has found any further result, please mention it in the comments.

#NumberTheory #RootsOfUnity #InfiniteSum #DivisorFunction

Easy Math Editor

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Comments

Consider writing $\displaystyle \sum_{b=1}^{\infty}{\sum_{a=1}^{\infty}{\frac{1}{a b (ab)!}}}$ and then probably your given identity(after some integration). Maybe! I haven't tried so I am not sure!

Kartik Sharma - 6 years, 1 month ago

I know how to transform it into the divisor sum, but I was wondering if we could find its exact value.Integration is not really a good idea, because tau is not a nice function to integrate.

Bogdan Simeonov - 6 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$