Is this series divergent?

Question

Is $$\sum_{n=1}^{\infty} \frac{\sqrt{n} - \left \lfloor {\sqrt {n}} \right \rfloor} {n} $$ divergent? I have tried both ratio test and root test, but they are definitely to weak to tackle this problem, I can't use integral test for this. I figured out that you can express this series as $$\sum_{k=1}^{\infty}\sum_{i=1}^{2k} \left( \frac{1}{\sqrt{k^2+i}} - \frac{k}{k^2+i}\right) $$

Answer 1

We have that

$$\sum_{n=1}^{N^2} \frac{\sqrt{n} - \left \lfloor {\sqrt {n}} \right \rfloor} {n}=\sum_{k=1}^{N}\sum_{n=(k-1)^2+1}^{k^2} \frac{\sqrt{n} - \left \lfloor {\sqrt {n}} \right \rfloor} {n}=\sum_{k=2}^{N}\left(-\frac1{k^2}+\sum_{n=(k-1)^2+1}^{k^2} \frac{\sqrt{n} -k+1} {n}\right)$$

and

$$\sum_{k=2}^{N}\left(\sum_{n=(k-1)^2+1}^{k^2} \frac{\sqrt{n} -k+1} {n}\right)\ge \sum_{k=2}^{N}\left( \frac{2k-1} {k^2}\right)$$

which diverges.

Answer 2

This is a short intuitive answer, note that fractional part of $\sqrt(n)$ are equidistributed $\mod 1$. Note that for each $n$ such that integral part of $sqrt(n)$ is $k$, there is about $k/2 +o(1)$ corresponding numbers whose square root has integral part $k$ and fractional part about $1/2$, hence you can convert the sum to about half of harmonic series. which diverges.