TITLE: Convergence of $a_{n}=\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{k}}$?
QUESTION [7 upvotes]: For $n$ in $\mathbb{N}$, consider the sequence $\left \{ a_{n} \right \}$ defined by:
$$a_{n}=\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{k}}$$
I would like to prove whether this sequence is convergent, and if so what its limit is.
I can prove by induction that $\sum\limits_{k=1}^{n}\frac{1}{\sqrt{k}}\geqslant \sqrt{n}$ For any $n$ in $\mathbb{N}$. Hence, $a_{n}\geqslant 1$. I wanted to prove that the sequence is decreasing and then use the monotone convergence theorem to prove it is convergent. However, I couldn't come up with a proof for this part. Anyone know how to prove convergence and find the limit?
I had another proof based on using Riemann sums, but I am looking for another proof using onne of the tricks used to prove convergence of sequences. Here is my proof: 
$$
a_{n}=\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=\sum_{k=1}^{n}\frac{1}{n}\frac{1}{\sqrt{\frac{k}{n}}}.
$$
Hence, $$\lim_{n \to \infty }a_{n}=\int_{0}^{1}\frac{dx}{\sqrt{x}}=2$$

REPLY [9 votes]: Using only elementary inequalities and no (improper) integral:
For every $k$, $\sqrt{k+1}-\sqrt{k}=\dfrac1{\sqrt{k+1}+\sqrt{k}}$ and $2\sqrt{k}\lt\sqrt{k+1}+\sqrt{k}\lt2\sqrt{k+1}$, hence
$$
\sqrt{k+1}-\sqrt{k}\lt\dfrac1{2\sqrt{k}}\lt\sqrt{k}-\sqrt{k-1}.
$$
Summing the rightmost inequality yields
$$
a_n\lt\frac2{\sqrt{n}}\sum_{k=1}^n\sqrt{k}-\sqrt{k-1}=2.
$$
Likewise, summing the leftmost inequality yields
$$
a_n\gt\frac2{\sqrt{n}}\sum_{k=1}^n\sqrt{k+1}-\sqrt{k}=2\frac{\sqrt{n+1}-1}{\sqrt{n}}\gt2-\frac2{\sqrt{n}}.
$$
Since $\dfrac2{\sqrt{n}}\to0$, this proves that $\lim\limits_{n\to\infty}a_n=2$.