# Does this sum of prime numbers converge?

$\newcommand{\P}{\operatorname{P}}$I'm wondering if this sum of prime numbers converges and how can I estimate the value of convergence.

$$\sum_{k=1}^\infty \frac{\P[k+1]-2\P[k+2]+\P[k+3]}{\P[k]-\P[k+1]+\P[k+2]}$$

$\$

\begin{align} \sum_{k=1}^{10} \frac{\P[k+1]\cdots}{\P[k]\cdots} & = 0.4380952380952381 \\ & \,\,\, \vdots \\ \sum_{k=1}^{10^5} \frac{\P[k+1]\cdots}{\P[k]\cdots} & =0.49433323447491884 \\[10pt] \sum_{k=1}^{10^6} \frac{\P[k+1]\cdots}{\P[k]\cdots} & = 0.49433634247938607`\ \approx \frac{5}{7}\zeta(3)^{-2} \end{align}

$\$

$\zeta(s) \$ is the Reimann zeta function

P$[n] \$ is the $n^\text{th}$ prime number

$\$

This is enough to assert that the series converges?

How can I estimate the value of convergence and if it is rational or irrational?

• How did you manage to conjecture $5/7 \zeta(3)^{-2}$ ? Jun 16 '17 at 17:03
• Where does this sum arise? Replacing $p_n$ with $n\ln n$ I believe the terms are bounded by $\frac 1{k\ln k}$ which would imply convergence. (Note I did it too quickly to be confident and one would need to check that it's ok to simply use the prime number theorem estimate.)
– lulu
Jun 16 '17 at 17:20
• @lulu I was looking at the gap between two prime numbers and I was wondering if it was to look for the link between the first number n and the number n + 1, it would be more predictable to take into account also the following n+2, n+3 numbers Jun 16 '17 at 17:42

As I commented on @user1952009's answer, the series converges under assumption of Cramer's conjecture. However, we can prove the convergence of the series unconditionally. We have in fact a result on partial sums of squares of prime gaps which was proven by R. Heath-Brown. Here's the link to the paper:

Theorem [Heath-Brown]

Let $p_n$ be the $n$-th prime, and let $g_n = p_{n+1}-p_n$. Then we have $$\sum_{n\leq x} g_n^2 \ll x^{\frac{23}{18}+\epsilon}.$$

After @user1952009's answer, it suffices to consider the convergence of (1): which is $$\sum_{k=2}^{\infty} \frac{g_k g_{k+c}}{k^2 \log^2 k} < \infty.$$

Note that $2g_kg_{k+c}\leq g_k^2 + g_{k+c}^2$, so it suffices to consider the convergence of $$\sum_{k=2}^{\infty} \frac{g_k^2}{k^2\log^2 k}$$ since the same idea will apply to $\sum_{k=2}^{\infty} \frac{g_{k+c}^2}{k^2\log^2 k}$. Let $A(x)=\sum_{k\leq x} g_k^2$, $f(x) = \frac1{x^2 \log^2 x}$. Heath-Brown's result states $A(x) \ll x^{23/18+\epsilon}$. Then by partial summation, we have \begin{align} \sum_{2\leq k\leq x} \frac{g_k^2}{k^2\log^2 k}&=\int_{2-}^x f(t) dA(t) \\ &=f(t)A(t) \bigg\vert_{2-}^x -\int_{2-}^x A(t) f'(t)dt\\ &=f(2-)A(2-) + O\left( x^{-\frac{13}{18}+\epsilon} \right)+\int_{2-}^x \frac{2A(t) (\log t + 1) }{t^3\log^3 t}dt \end{align} Now, we have the convergence since $23/18 - 3 = -31/18<-1$.

Another problem that uses Heath-Brown's result is here.

Letting $b_k = p_k - p_{k+1}+p_{k+2}$ and $a_k = g_{k+2} - g_{k+1}$ where $g_k = p_{k+1}-p_k$,

summing by parts $$\sum_{k=1}^n \frac{a_k}{b_k} = \frac{g_2-g_{n+2}}{b_n}+\sum_{k=1}^{n-1} (g_2-g_{k+2})(\frac{1}{b_k}- \frac{1}{b_{k+1}})$$ Since $b_k \sim k \log k$ $$\frac{1}{b_k}- \frac{1}{b_{k+1}}= \frac{b_k-b_{k+1}}{b_kb_{k+1}}\sim \frac{g_k+g_{k+1}+g_{k+2}}{k^2 \log^2 k}$$ Also we know $g_k = \mathcal{O}(k^\theta)$ for some $\theta < 0.6$ but this is not sufficient to conclude.

We look instead at $$\sum_{k=2}^\infty \frac{g_k g_{k+c}}{k^2 \log^2 k} \overset{?}< \infty \tag{1}$$

By summation by parts $\sum_{k=2}^K \frac{g_{k+c}}{k \log k} \sim \log K$ so I'd say yes $\sum_{k=2}^\infty \frac{g_k g_{k+c}}{k^2 \log^2 k}$ converges, but I can't prove it.

Summing by parts $\sum_{k=K}^\infty \frac{g_k }{k^2 \log^2 k} \sim \frac{1}{k \log k}$ lets me think $(1)$ converges.

• I appreciate your answer, it is convincing. I do not have enough knowledge to say that it is true. Now the value for which the series converges and whether this is rational or irrational remains to be calculated Jun 16 '17 at 20:33
• @PatrickDanzi I made a mistake, see the edits Jun 16 '17 at 21:42
• Cramer's conjecture would imply convergence. Jun 17 '17 at 2:54
• Nice (+1). This idea leads to an unconditional proof of convergence. Jun 17 '17 at 3:40