Proving that $\sum_p\frac{1}{p+1}$ diverges

Question

How does one prove $$\sum_{p\in\Bbb P}\frac1{p+1}=\infty.$$

Where $\Bbb P$ denotes the set of prime numbers.

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I have attempted forming an inequality by playing around with Euler's work on the reciprocals of primes. Robjohn showed me an inequality in chat that I do not understand and I was wondering if there was another way to do this. The work I have done is inconclusive and is nothing more than stating $$\sum_{p\in\Bbb P}\frac1p\ge\sum_{p\in\Bbb P}\frac1{p+1}$$

Answer 1

$$ \sum \frac{1}{p+p} < \sum \frac{1}{p+1} \implies \frac{1}{2} \sum \frac{1}{p} < \sum \frac{1}{p+1}. $$

Answer 2

You might want to use that $p+1$ is less than the next prime after $p$ except when $p=2$. So you can use a comparison, and drop the first term, which doesn't affect convergence/ divergence.