Does the sum $\sum_{n \geq 1} \frac{2^n\operatorname{mod} n}{n^2}$ converge? I am somewhat a noob, and I don't recall my math preparation from college. I know that the sum $\displaystyle \sum_{n\geq 1}\frac{1}{n}$ is divergent and my question is if the sum$$\sum \limits _{n\geq 1}\frac{2^n\mod n}{n^2}$$converges. I think is not but I do not know how to prove that! Thanks!

Not an answer, just for insight. Here is a plot of the $1000$ first $\dfrac{2^n\mod n}n$, to substantiate the "uniform distribution" hypothesis.

The second plot is the prefix sum, which is a straight line for a uniform distribution. The average of the values is $0.287$ (not very close to $0.5$).

Unfortunately, this empirical data isn't sufficiently conclusive.

Here are histograms of the fractional part of the $10000$, $100000$ and $1000000$ first values. Beware that the frequency axis is logarithmic. There is a strong bias towards the small values (one third below $0.01$), but this doesn't seem to worsen with $n$.



I conjecture that for any $n$, at least $50\%$ of the values of $\{2^k/k\}$ are above $0.1$. 
Very interestingly, the histogram of the values $2^n\bmod n$ themselves shows very strong lines for all powers of $2$.

This leads us to the next histogram, where the moduli that are powers of $2$ have been ignored. There is no more need for a logarithmic scale and the distribution looks pretty flat now.

