# Convergence of $\sum\limits_{k=1}^{\infty} \frac{1}{p_{k^2}}$, where $p_k$ is the $k$th prime

Two brief questions. This seems true but I don't find it using Google. (1) Isn't

$$\sum_{k=1}^{\infty} \frac{1}{p(k^2)}$$, in which $p(k^2)$ is the $k^2$th prime,

known to converge? I expected to find something in Plouffe around 0.747187, but did not.

It seems that $p(k^2) > k^2$ proves convergence, but then did I miscalculate the constant?

It occurred to me that we might we use comparison (via the PNT) for a series, but all we know is that termwise $$\frac{1}{p(k^2)}\sim \frac{1}{k^2\ln k^2},$$ and I don't think that $a\sim b$ and $c \sim d$ gives $a+c \sim b+d$ ? So, (2) is it correct that the logarithmic sum also converges but gives no clue as to convergence of the sum in question?

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We certainly don't need the Prime Number Theorem, or indeed anything, to show that the $k^2$-th prime is $\ge k^2$. –  André Nicolas Dec 13 '11 at 3:05
Of course not. But I wondered whether my rejection of the possibility was correct. Question (2). –  daniel Dec 13 '11 at 3:08
But then you know that the series converges. So the question cannot be about convergence. –  André Nicolas Dec 13 '11 at 3:12
After much effort (SequenceLimit[Drop[N[Accumulate[1/Prime[Range[7000]^2]], 600], 2000]]), the best estimate I have is 0.7471881931, but I'd only trust the first three digits... –  Guess who it is. Dec 13 '11 at 5:27
Let $f(n) = \sum_{k=1}^n 1/p(k^2) + \int_n^\infty 1/((n \log n)^2)$. (The latter integral, Maple tells me, can be written in terms of elliptic integrals.) Evaluating $f(2^2), f(2^3), \ldots, f(2^{10})$ and doing Richardson extrapolation (en.wikipedia.org/wiki/Richardson_extrapolation) on the sequence gives an estimate of $0.7471910318$. I'm inclined to say I trust this to seven places but that may be overoptimistic. –  Michael Lugo Dec 13 '11 at 22:03

It seems to me that most of the questions have been settled in the comments. I'll try to pull it all together.

Yes, the series converges, most simply by comparison with $\sum k^{-2}$.

No, you didn't miscalculate the constant. Well, I can't prove that, but J. M. corroborates your calculation, and that will do me.

No, the constant doesn't seem to appear anywhere. As to why, I can only speculate that no one finds the sequence 1st prime, 4th prime, 9th prime, 16th prime, etc., especially interesting, so no one has been motivated to do any calculations related to it. Indeed, you haven't given us any reason to be interested in it, any connection with any other part of mathematics.

Yes, $\sum(k^2\log(k^2))^{-1}$ converges, and its convergence can be used to prove that your sum converges, and can probably be used to give some bound on your sum. It certainly won't help you find the exact value of your sum.

Have I left anything out?

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Steven R Finch, Mathematical Constants, in the section on Meissel-Mertens constants, writes, "The sum of the squared reciprocals of primes is $$N=\sum_p{1\over p^2}=\sum_{k=1}^{\infty}{\mu(k)\over k}\log(\zeta(2k))=0.4522474200\dots"$$ and connects it to the variance of $\omega(n)$ (which is the number of distinct prime divisors of $n$).

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I think $\sum\limits_k\frac1{p_k^2}$ and $\sum\limits_k\frac1{p_{k^2}}$ are entirely different things... –  Guess who it is. Dec 13 '11 at 4:48
How different they are would depend on how good your eyesight is. I'll leave my "answer" up as a warning to everyone else. –  Gerry Myerson Dec 13 '11 at 4:53