How to prove that $\sum _{p}\sum_{k = 2}^\infty \frac{1}{kp^k}$ is convergent?

Question

How to prove that $\sum _{p}\sum_{k=2}^\infty \frac{1}{kp^k}$ is convergent? where $p$ is prime.

What I have done is $$\sum_{k = 2}^\infty \frac{1}{kp^k}=-\log\left(1-\frac 1p\right)-\frac 1p $$

Then for $\sum _{p}(-\log(1-\frac 1p)-\frac 1p) $ I was trying to use integral test after putting $n\in \Bbb N$ in place of $p$. But I was not getting my desired result. Please help.

Answer 1

For any $x\in\left[0,\frac{1}{2}\right]$ we have$^{(*)}$ $0\leq -\log(1-x)-x \leq x^2$, hence your series is convergent by comparison with $\sum_{p}\frac{1}{p^2}.$ Proof of $(*)$: it is enough to show $$ x \leq -\log(1-x) \leq x+x^2\qquad \forall x\in\left[0,\tfrac{1}{2}\right] $$ or $$ 1\leq \frac{1}{1-z}\leq 1+2z\qquad \forall z\in\left[0,\tfrac{1}{2}\right] $$ then integrate such terms over $[0,x]$. The last inequality is equivalent to $$ 1-z \leq 1 \leq 1+2z\left(\tfrac{1}{2}-z\right)\qquad \forall z\in\left[0,\tfrac{1}{2}\right] $$ which is trivial.

Answer 2

Use Taylor series of log function: $$ \log\left(1-x\right) = -(x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots) = -x+O(x^{2}) $$