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.

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TopNewestConsider 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 · 2 years, 2 months ago

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– Bogdan Simeonov · 2 years, 2 months 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.Log in to reply