Looking for a tight upper bound of a sum

Question

I try to upper bound tightly the following sum:

$$\sum_{i=1}^N \left(\frac{1}{\sqrt{i}} - \frac{1}{\sqrt{i+1}}\right)i$$

I expect something which is asymptotically close to $\sqrt{N}$.

Thank you very much.

Answer 1

By creative telescoping,

$$\begin{eqnarray*} \sum_{i=1}^{N}\left(\frac{1}{\sqrt{i}}-\frac{1}{\sqrt{i+1}}\right)i =\sum_{i=1}^{N}\frac{\sqrt{i}}{\sqrt{i+1}(\sqrt{i}+\sqrt{i+1})}&\color{red}{\leq}&\sum_{i=1}^{N}\frac{1}{\sqrt{i}+\sqrt{i+1}}\\&=&\sum_{i=1}^{N}\left(\sqrt{i+1}-\sqrt{i}\right)\\&=&\sqrt{N+1}-1.\end{eqnarray*}$$ More carefully, $$\begin{eqnarray*} S_N &=& \sum_{i=1}^{N}\sqrt{i}-\sum_{i=1}^{N}\sqrt{i+1}+\sum_{i=1}^{N}\frac{1}{\sqrt{i+1}}\\&=&-\sqrt{N+1}+H^{(1/2)}_{N+1}\\&\leq &\sqrt{N+1}+\zeta\left(\tfrac{1}{2}\right)+\frac{1}{2\sqrt{N+1}}\end{eqnarray*}$$ where $\zeta\left(\tfrac{1}{2}\right)=-1.4603545088\ldots$.