Is $\sum_{n=1}^{+ \infty}\left(\frac {1}{n}-\frac{1}{p_n}\right)$ convergent?

Question

It is known that $$\sum_{n=1}^{+ \infty} \frac {1}{n}$$ is divergent. Also, it is known that $$\sum_{n=1}^{+ \infty} \frac {1}{p_n}$$ is divergent where $p_n$ is $n$-th prime number.

I was thinking what would happen (in the sense of convergence) if we termwise subtract these two series to obtain $$\sum_{n=1}^{+ \infty} \left(\frac {1}{n}-\frac{1}{p_n}\right)$$

Is $$\sum_{n=1}^{+ \infty} \left(\frac {1}{n}-\frac{1}{p_n}\right)$$ convergent?

Answer 1

Here's an easy answer that doesn't require heuristics or (much) number theory: first of all, note that $p_n\gt 2n$ for $n\gt 4$. (Can you see why?) Now, this gives $\frac1{p_n}\lt\frac1{2n}$, and therefore $\frac1n-\frac1{p_n}\gt\frac1n-\frac1{2n} = \frac1{2n}$. Now just use the comparison test against the harmonic series $\sum_n\frac1{2n}$.

Answer 2

Using the estimate $p_n \approx n\ln(n)$, we get

$\begin{array}\\ \sum_{n=1}^{m}( \frac {1}{n}-\frac {1}{p_n}) &=\sum_{n=1}^{m}\frac {1}{n}-\sum_{n=1}^{m}\frac {1}{p_n}\\ &\approx \ln(m)+O(1)-\sum_{n=1}^{m}\frac {1}{n \ln(n)}\\ &\approx \ln(m)+O(1)-\ln\ln(m)+O(1)\\ &= \ln(m)-\ln\ln(m)+O(1)\\ &\to \infty\\ \end{array} $