TITLE: Evaluate the limit $\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)$
QUESTION [17 upvotes]: Evaluate the limit $\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)$, without using a Riemann sum

$\bf{My\; Try:}$ Using the graph of $\displaystyle f(x) = \frac{1}{\sqrt{x}}\;,$ we get
$$\int_{1}^{n+1}\frac{1}{\sqrt{x}}dx <\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)<1+\int_{1}^{n}\frac{1}{\sqrt{x}}dx$$
So we get $$2\sqrt{n+1}-2<\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)<2\sqrt{n}-1$$
So $$\lim_{n\rightarrow \infty}\frac{2\sqrt{n+1}-2}{\sqrt{n}}=2<\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)<\lim_{n\rightarrow \infty}\frac{2\sqrt{n}-1}{\sqrt{n}} = 2$$
So using the Sandwich Theorem, we get $\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)=2$
My question is can we solve it by using any other method? If yes then please explain it here.

REPLY [4 votes]: Notice that:
$$\frac1{\sqrt n}\sum_{k=1}^n\frac1{\sqrt k}=\frac1n\sum_{k=1}^n(k/n)^{-1/2}$$
As $n\to\infty$, we get a Riemann sum:
$$\lim_{n\to\infty}\frac1n\sum_{k=1}^n(k/n)^{-1/2}=\int_0^1x^{-1/2}~\mathrm dx=2x^{1/2}\bigg|_0^1=2$$