I want to find an analytical expression for:
$\frac{1}{n}\left[1+2\sum_{k=1}^{n-1}\frac{1}{\sqrt{\frac{n}{n-k}}}\right]$
I know that the result is independent of $n$ when $n$ is large, because I have used MATLAB for many different values of $n$, and the answer always comes out as 4/3, but I am thus far unable to rearrange or re-express this summation to be able to show this analytically.
Any suggestion would be much appreciated. This is not for any kind of assessed work.
$$\frac{1}{n}\left[1+2\sum_{k=1}^{n-1}\frac{1}{\sqrt{\frac{n}{n-k}}}\right]=\frac{1}{n}+2\frac{1}{n}\sum_{k=1}^{n-1}\frac{\sqrt{n-k}}{\sqrt{n}}=\frac{1}{n}+2\frac{1}{n}\sum_{k=1}^{n-1}\sqrt\frac{n-k}{n}$$= $$\frac{1}{n}+2\frac{1}{n}\sum_{k=1}^{n-1}\sqrt{1-\frac{k}{n}}$$. When $ n \to \infty$, this turns out to be $$2\int_0^1 \left(\sqrt{1-x} \right)dx$$
Hint :
You can rewrite $\displaystyle\frac{1}{n}\sum_{k=1}^{n-1} \frac{1}{\sqrt{\frac{n}{n-k}}}$ as $$ \frac{1}{n} \sum_{k=1}^{n-1} \sqrt{1-\frac{k}{n}} $$ and then use Riemann sum for an appropriate function.
The first term $1/n\to 0$, and the rest is a Riemann sum for the integral $2\int_0^1 \sqrt{1-x} \, dx = 4/3$, so that's what it converges to as $n \to \infty$.
