I am trying to compute the following infinite sum

$$\sum_{n=2}^\infty \,\frac{\Lambda(n)}{n^2 \ln n},$$

where $\Lambda(n)$ is Mangoldt's function. It seems to me that the result is strictly less than 1/2. Does somebody know a reference where that series is computed?

  • 2
    $\begingroup$ You probably already noticed that, but it is easy to see that the series converges and to give an upper bound by noticing that $$\sum_{n=2}^\infty\frac{\Lambda(n)}{n^2\log n}<\sum_{n=2}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}-1\approx0.644934\ .$$ Unfortunately, it is not the wanted $\frac{1}{2}$, and naively considering a few more terms don't give us such a strong bound. $\endgroup$ Sep 10, 2017 at 16:25
  • 3
    $\begingroup$ $$\sum _{n=2}^{\infty } \frac{\Lambda (n)}{n^2 \log (n)}=\log (\zeta (2))\approx 0.4977003024707452$$ en.wikipedia.org/wiki/Von_Mangoldt_function#Dirichlet_series $\endgroup$
    – Raffaele
    Sep 10, 2017 at 16:47

2 Answers 2


$$\log \zeta(s) = \log \prod_p \frac{1}{1-p^{-s}} = \sum_p - \log(1-p^{-s}) = \sum_p \sum_{k=1}^\infty \frac{p^{-sk}}{k}$$

$$\sum_{n=2}^\infty \frac{\Lambda(n)}{ \log n}n^{-2} = \sum_{p}\sum_{k =1}^\infty \frac{p^{-2k}}{k} = \log \zeta(2) = 2 \log \pi - \log 6$$

  • $\begingroup$ Of course, one can evaluate the double sum in terms of the log of the zeta function. I totally missed this, even though the initial spirit of our answers is the same. Excellent work. $\endgroup$
    – davidlowryduda
    Sep 10, 2017 at 16:48

One way to approach this sum is to cast it in terms of the prime zeta function, $$ P(s) = \sum_{p \text{ prime}} \frac{1}{p^s}.$$ Then $$ \sum_{n \geq 2} \frac{\Lambda(n)}{n^2 \log n} = \sum_{p} \sum_{k \geq 1} \frac{\Lambda(p^k)}{p^{2k} \log (p^k)} = \sum_{k \geq 1} \frac{1}{k} \sum_{p} \frac{1}{p^{2k}} = \sum_{k \geq 1} \frac{1}{k} P(2k). \tag{1}$$ We can estimate $(1)$ by summing the first several terms and bounding the remainder. This leads us to consider $$ \sum_{1 \leq k \leq K-1} \frac{1}{k} P(2k) + \sum_{k \geq K} \frac{1}{k} P(2k).$$ Taking the first 2000 terms shows the first sum to be about $0.49770030\ldots$ Let us now focus entirely on the remainder.

Note the loose trivial bound $$ P(n) \leq \big( \zeta(n) - 1\ \big).$$ The $-1$ has the simple effect of removing the leading $1$ from the zeta function, so that we are left with only rapidly decreasing terms. Then we have $$ \sum_{k \geq K} \frac{1}{k} P(2k) \leq \sum_{k \geq K} \frac{1}{k} \big(\zeta(2k) - 1\big).$$ Using the integral-test-inequality $$ \sum_{n \geq 2} \frac{1}{n^{2k}} \leq \int_2^\infty \frac{1}{t^{2k}}dt + \frac{1}{2^{2k}} = \frac{1}{2k-1} \frac{1}{2^{2k-1}} + \frac{1}{2^{2k}},$$ we bound the error by $$ \sum_{k \geq K} \frac{1}{k} \bigg( \frac{1}{2k-1} \frac{1}{2^{2k-1}} + \frac{1}{2^{2k}}\bigg),$$ for which it's very easy to see that this doesn't contribute meaningfully to the main term above (and decreases exponentially in $K$).

So the infinite series equals $0.49770030\ldots$


You must log in to answer this question.