# Does converge $\sum_{n=2}^\infty\frac{1}{\varphi(p_n-2)-1+p_n}$, where $\varphi(n)$ is the Euler's totient function and $p_n$ the $n$th prime number?

I was thinking this exercise, and now I don't know if my series is obviously convergent (I was writting simple manipulations from the AM–GM inequality).

The series is $$\sum_{n=2}^\infty\frac{1}{\varphi(p_n-2)+\varphi(p_n)}=\sum_{n=2}^\infty\frac{1}{\varphi(p_n-2)-1+p_n},\tag{1}$$ where here $p_n$ denotes the $n$th prime number and $\varphi(m)$ the Euler's totient function.

I think that doesn't converge.

Question. Do you know how deduce the asymptotic behaviour of $$\sum_{2\leq n\leq x}\frac{1}{\varphi(p_n-2)+\varphi(p_n)},\tag{2}$$ as $x\to\infty$? Or, deduce if our series is convergent Thanks you in advance.

You can use that for $n \geq 3$ $$\varphi\left(p_n-2\right) \leq p_n-3 \text{ and }\varphi\left(p_n\right)=p_n-1$$ So $$\varphi\left(p_n-2\right)+\varphi\left(p_n\right) \leq p_n-3+p_n-1= 2p_n-4$$ Then consider the reverse expression for $n \geq 3$

$$\frac{1}{\varphi\left(p_n-2\right)+\varphi\left(p_n\right)} \geq \frac{1}{2p_n-4}$$ Furthermore for $n \geq 3$ it is strictly positive and $$\frac{1}{2p_n-4} \underset{(+\infty)}{\sim}\frac{1}{2p_n}$$ And the series $\displaystyle \sum_{n \geq 1}^{ }\frac{1}{p_n}$ diverges, so as your series.

• Yes very intelligent. I understand your proof, many thanks.
– user243301
Jan 18 '18 at 17:18