On a conjecture that $\sum\limits_{n=1}^k\frac{1}{\pi^{1/n}p_n}\stackrel{k\to\infty}{\longrightarrow} 2$. I have made the following conjecture, and I do not know if this is true.


Conjecture:

\begin{equation*}\sum_{n=1}^k\frac{1}{\pi^{1/n}p_n}\stackrel{k\to\infty}{\longrightarrow}2\verb| such that we denote by | p_n\verb| the | n^\text{th} \verb| prime.|\end{equation*}

Is my conjecture true? It seems like it, according to a plot made by Wolfram|Alpha, but if it does, then it converges.... very.... very, slowly. In fact, let $k=5000$, then the sum is approximately equal to $1.97$, which just proves how slow it would be.
Is there a way of showing whether or not this is indeed convergent? For any other higher values of $k$, it seems that it is just too much for Wolfram|Alpha to calculate, and it does not give me a result when I let $k=\infty$. Also, for users who might not understand the notation, we can similarly write that $$\sum_{n=1}^\infty\frac{1}{\pi^{1/n}p_n}=2\qquad\text{ or }\qquad\lim_{k\to\infty}\sum_{n=1}^k\frac{1}{\pi^{1/n}p_n}=2.$$ Also, without Wolfram|Alpha, I have no idea how to approach this problem in terms of proving it or disproving it. Does the sum even converge at all? If so, to what value? Any help would be much appreciated.

Thank you in advance.
Edit:
I looked at this post to see if I could rewrite my conjecture as something else in order to help myself out. Consequently, I wrote that $$\sum_{n=1}^k\frac{1}{\pi^{1/n}p_n}\stackrel{k\to\infty}{\longleftrightarrow}4\sum_{n=1}^\infty\frac{1}{n^k+1}\tag{$\text{LHS}=2$}$$ since both sums look very similar. Could this be of use?
 A: Recall that
$\sum_{n=1}^{\infty} \frac{1}{p_n}$ is a divergent series. Then your series is divergent too because, for any positive number $a$,
$$\lim_{n\to \infty}a^{1/n}=\lim_{n\to \infty}e^{\ln(a)/n}=1,$$ and therefore
$$\frac{1}{\pi^{1/n}p_n}\sim \frac{1}{p_n}.$$
A: Numerically, this does not seem to be true.
Considering $$S_m=\sum_{n=1}^{10^m}\frac{1}{\pi^{1/n}p_n}$$ and, using illimited precision, I obtained the following numbers
$$\left(
\begin{array}{cc}
 m & S_m \\
 1 & 0.891549393 \\
 2 & 1.437754209 \\
 3 & 1.787152452 \\
 4 & 2.038881140 \\
 5 & 2.235759176 \\
 6 & 2.397832041
\end{array}
\right)$$
Edit
After marty cohen's answer, based on the above data, a quick and dirty fit for the model
$$S_k=\sum_{n=1}^{k}\frac{1}{\pi^{1/n}p_n}=a+b\,\log(\log(k))$$ gives $(R^2=0.999947)$
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 a & 0.17085 & 0.02635 & \{0.08600,0.25471\} \\
 b & 0.84291 & 0.01302 & \{0.80146,0.88436\} \\
\end{array}$$
A: $\sum_{n=1}^{\infty} \frac{1}{s_n}$is a divergent series therefore if you tend the limit to $\infty$ ;
$\lim_{n\to \infty}\pi^{1/n}=1$
You can see that the above term tends to $1$.
The remaining term is similar to above mentioned divergent series. 
A: If $a > 1$
Then,
since
$p_n \sim n \ln n$
and
$1/a^{1/n}
=e^{-\ln a/n}
\sim 1-\ln a/n
$,
$\sum_{k=1}^n 1/(a^{1/k}p_k)
\sim \sum_{k=1}^n 1/p_k-\sum_{k=1}^n\ln a/(kp_k)
\sim \ln. \ln n-c
$
for some $c$
since the second sum converges.
Therefore the sum diverges like
$\ln \ln n$.
