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 $k\ge 3$, $k\in\mathbb Z.$
 A: 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.
A: Throwing in all integers greater than $1$ still gives a convergent series:
$$
\begin{align}
\sum_{k=2}^\infty\sum_{p\in\mathbb{P}}\frac1{k\,p^k}
&\le\sum_{k=2}^\infty\sum_{n=2}^\infty\frac1{k\,n^k}\\
&=\sum_{n=2}^\infty\left[\log\left(\frac{n}{n-1}\right)-\frac1n\right]\\
&=\lim_{m\to\infty}\sum_{n=2}^m\left[\log\left(\frac{n}{n-1}\right)-\frac1n\right]\\[3pt]
&=\lim_{m\to\infty}\left[\log(m)-H_m+1\right]\\[9pt]
&=1-\gamma
\end{align}
$$
A: 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.
A: Since the terms are positive we can rearrange them
$$K:=\sum_{k=2}^{\infty}\frac{1}{k}\sum_{p\in\mathbb{P}}\frac{1}{p^k}=
\sum_{p\in\mathbb{P}}\sum_{k=2}^{\infty}\frac{(1/p)^k}{k}=\sum_{p\in\mathbb{P}}\left[-\ln\left(1-(1/p)\right)-(1/p)\right].$$
Now note that $-\ln(1-x)-x\leq x^2$ for $x\in [0,1/2]$ (because the derivative of the difference is $\frac{x(1-2x)}{1-x}\geq 0$) which implies that
$$K\leq \sum_{p\in\mathbb{P}}\frac{1}{p^2}\leq \sum_{k=2}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}-1<1.$$
