Prove $\frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{3}\sum_p \frac{1}{p^3}+\cdots$ converges 
Prove $$\frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{3}\sum_p \frac{1}{p^3}+\cdots$$ converges, where the sums are for all primes $p$.

I've found in this link that $$\frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{3}\sum_p \frac{1}{p^3}+\cdots$$
converges to $K$, where $K<1$.
I know that $\sum_p \frac{1}{p^2}\le \sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6},$
because see Basel problem and specific values of Riemann zeta function.
Therefore also $$\sum_p \frac{1}{p^k}\le \sum_p \frac{1}{p^2}\le\sum_{n=1}^{+\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$
for all $k\ge 3$, $k\in\mathbb Z.$

Notice that
$$\sum_{p} \frac{1}{p^s}\lt\zeta(s)-1$$
so your sum is bounded from above by
$$
\begin{align}
\sum_{s=2}^\infty(\zeta(s)-1)
&=\sum_{s=2}^\infty \sum_{n=2}^\infty \frac{1}{n^s}\\
&=\sum_{n=2}^\infty \sum_{s=2}^\infty \frac{1}{n^s}\\
&=\sum_{n=2}^\infty\left(\frac{1}{1-\frac{1}{n}}-1-\frac{1}{n}\right)\\
&=\sum_{n=2}^\infty\left(\frac{1}{n-1}-\frac{1}{n}\right)\\[9pt]
&=1
\end{align}
$$
and so your sum is less than one, and must also trivially be positive, so it converges.
