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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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    The complete asymptotic expansion is computed at this [MSE link](http://math.stackexchange.com/questions/79115/limit-lim-limits-n-rightarrow-infty-left2-sqrt-n-sum-limits-k-1n-frac).2014-03-30

2 Answers 2

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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 $\,2$ .