# Convergence of $a_{n}=\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{k}}$?

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, $$\lim_{n \to \infty }a_{n}=\int_{0}^{1}\frac{dx}{\sqrt{x}}=2$$

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This is the second time you've proposed to use one of the powerful and more-or-less advanced convergence theorems: the monotone one, the dominated one...why? Have you people just studied them now? Because the exercises you've posted are way more elementary and hardly require using those heavy weapons... –  DonAntonio Jul 13 '12 at 16:17
I think by the monotone convergence theorem, OP doesn't mean the one in integration theory, but just the argument that a monotone, bounded sequence converges, which is straightforward from the definition of $\mathbb{R}$. –  user29743 Jul 13 '12 at 18:26
In fact the sequence is increasing, not decreasing. –  Robert Israel Jul 13 '12 at 19:15
The complete asymptotic expansion is computed at this MSE link. –  Marko Riedel Mar 30 at 19:53

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 $\lim\limits_{n\to\infty}a_n=2$.

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$$\frac{1}{\sqrt n}\,\sum_{k=1}^n\frac{1}{\sqrt k}=\sum_{k=1}^n\frac{1}{n}\sqrt{\frac{n}{k}}\xrightarrow [n\to\infty]{} \int_0^1\frac{dx}{\sqrt x}=\lim_{\varepsilon\to 0^+}2\left(\sqrt 1-\sqrt \varepsilon\right)$$
The above is an improper integral (not bounded by one of the limit points), but it's easy to check it converges to $\,2$ .