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 $k\ge 3$, $k\in\mathbb Z.$

Since $2$ is the smallest prime, we can see by comparing terms that
$$\frac{1}{2}\sum_p \frac{1}{p^2}>\sum_p \frac{1}{p^3}$$
$$\frac{1}{2}\sum_p \frac{1}{p^3}>\sum_p \frac{1}{p^4}$$
and so on.
So it follows that 
$$\frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{3}\sum_p \frac{1}{p^3}+\cdots < \frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{2}\sum_p \frac{1}{p^3}+\cdots <$$
$$\frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{4}\sum_p \frac{1}{p^2}+\cdots =\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots\right)\sum_p\frac{1}{p^2}=$$
$$\sum_p \frac{1}{p^2}<\sum_n \frac{1}{n^2} =\frac{\pi^2}{6}$$
Since the sum is increasing, the presence of an upper bound means it converges.
