Edit Motivation for question: I looked up the decimal expansion of: $$\sum _{n=1}^{\infty } \sum _{k=n}^{\infty } k^{-2 k},$$ which matches the first seven digits of the function in question. I would like to investigate more, but I don't know where to look.

I found this series at OEIS A096250 $$\sum_{n=1}^{\infty}n^{-p_{n}}$$

Can someone point me to a reference?
What is the significence of this series? Does it say anything about the distribution of the primes, for instance?

  • $\begingroup$ Do you expect anything about the distribution of squares from $\sum_{n=1}^{\infty}n^{-n^2}$? $\endgroup$
    – draks ...
    Commented May 1, 2013 at 14:34
  • $\begingroup$ @draks... I don't know. When I first saw it, it didn't seem to make sense. Now, I not sure. Anyway, I'd like a reference, if any are known. $\endgroup$ Commented May 1, 2013 at 15:21

1 Answer 1


Rearanging your series, we have that

$$\sum_{n=1}^{\infty}\sum_{k=n}^{\infty}\frac{1}{k^{2k}}=\sum_{k=1}^{\infty}\frac{1}{k^{2k}}\sum_{n=1}^{k}1=\sum_{k=1}^{\infty}\frac{1}{k^{2k-1}}.$$ Now, you are asking why is the above so close to $$\sum_{n=1}^{\infty}\frac{1}{n^{p_{n}}},$$ where $p_{n}$ is the $n^{th}$ prime number. The reason is the following: The terms in these series decrease to $0$ extremely quickly, faster than the function $n^{-n}$, and since the first few terms are the same for both series, it follows that the error will be very small. Notice that $$\sum_{n=1}^{\infty}\frac{1}{n^{p_{n}}}=1+\frac{1}{2^{3}}+\frac{1}{3^{5}}+\frac{1}{4^{7}}+\frac{1}{5^{11}}+\cdots,$$ whereas $$\sum_{k=1}^{\infty}\frac{1}{k^{2k-1}}=1+\frac{1}{2^{3}}+\frac{1}{3^{5}}+\frac{1}{4^{7}}+\frac{1}{5^{9}}+\cdots,$$ so the difference between them is on the order of $$\frac{1}{5^{9}}-\frac{1}{5^{11}}=4.9152\times10^{-7},$$ which explains why there agree for the first $7$ digits. If you are curious why the first few terms agree, it is because each of $3,5,7$ can be written as $2k-1$, for $k=2,3,4$.


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