# Shifted sum of reciprocals of primes: $\sum \limits_{\forall p \in primes}\frac{1}{p+1}$

I know that the sum:

$\sum \limits_{\forall p \in primes}\frac{1}{p}$ diverges.

I was wondering if:

$\sum \limits_{\forall p \in primes}\frac{1}{p+1}$ diverges as well.

And if there is a formula for the partial sum of primes up to an upperbound $n$.

My guess is that it does but I'm not quite sure what the quick way to show it is. Any help is greatly appreciated!

• Hint: limit comparison test. – Mike Earnest Jun 1 '18 at 14:57
• The other solutions (and hints) are fine but you could also just note that $p_n+1≤p_{n+1}$ so you can directly compare the two. (Note: here $p_n$ denotes the $n^{th}$ prime.) – lulu Jun 1 '18 at 14:59

Let $p_n$ be the $n$-th prime. Then these two series diverge:
$$\sum_{n=1}^{\infty} 1/p_n$$ $$\sum_{n=1}^{\infty} 1/p_{n+1}$$ Since $p_{n+1} \geq p_n + 1$, we conclude that the series you described is divergent as well.
Hint: $$\frac{1}{p+1} \ge \frac{1}{2p}.$$ For partial sum of reciprocals of primes asymptotic is well known: $$\sum_{p< n} \frac{1}{p} = \log \log n + M_n,$$ where $\lim_{n\to \infty}M_n = M$ is Meissel–Mertens constant.
• The last equation is not correct. The $M$ should be replaced by $M_n$ where $\lim_{n\to \infty}M_n=M.$ – DanielWainfleet Jun 1 '18 at 17:30