# 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.

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?

• Hint: $p_n \sim n \log n$. Commented May 30, 2018 at 5:24
• @JavaMan could you please elaborate? Thanks. Commented May 30, 2018 at 5:27
• For the numbers I used, $S_m\approx \sqrt m$. Commented May 30, 2018 at 6:16

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 $$\frac{1}{\pi^{1/n}p_n}\sim \frac{1}{p_n}.$$

• I never knew that the sum of the reciprocal of primes was a divergent series. That makes Brun's Constant even more special. Congratulations! $$(+1) \ \ \color{green}{\checkmark}$$ Commented May 31, 2018 at 0:44

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 $(R^2=0.999947)$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 0.17085 & 0.02635 & \{0.08600,0.25471\} \\ b & 0.84291 & 0.01302 & \{0.80146,0.88436\} \\ \end{array}$$

• It appears that by letting $m_n > m_{n-1}$, we have that $m_n-m_{n-1}\to0$. Commented May 31, 2018 at 0:42

$\sum_{n=1}^{\infty} \frac{1}{s_n}$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.

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 $\ln \ln n$.