# Limit of $s_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)$

\begin{align*}S_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)\end{align*}

how to calculate the limit $s_n$?

\begin{align*}\lim_{n\to \infty } \, S_n\end{align*}

• What is the question you are trying to ask? What thoughts do you have about the problem you are trying to address? Please edit your post heavily to put it into proper question form (including your own input on the problem and issues you may be facing). – Cameron Williams Sep 16 '13 at 2:52
• $\displaystyle{\Large 2\ ?}$. – Felix Marin Sep 16 '13 at 3:35
• Can one add an answer via using stolz? – polynomial Sep 16 '13 at 3:37
• @polynomial If you prefer an answer based on some specific method, I'll suggest you include that preference into your question. – achille hui Sep 16 '13 at 3:45
• The Maple command $$limit((sum(1/sqrt(k), k = 1 .. n))/sqrt(n), n = infinity)$$ outputs 2. – user64494 Sep 16 '13 at 4:36

Consider the curve $y=\frac{1}{\sqrt{x}}$. We have $$\int_1^{n+1}\frac{1}{\sqrt{x}}\,dx\lt 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}\lt \int_0^n \frac{1}{\sqrt{x}}\,dx.$$ Evaluate the integrals. We get $$2\sqrt{n+1}-2\lt 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}\lt 2\sqrt{n}.$$ Divide everything by $\sqrt{n}$, and use Squeezing to conclude that our limit is $2$.

• Thanks. I have trouble with constants, basically pay no attention to them. Works well, except when it doesn't. – André Nicolas Sep 16 '13 at 3:28
• You must be a measure theorist ;) – jwg Sep 16 '13 at 6:43
• No, just have taught too many calculus courses. But I am estimation-inclined. – André Nicolas Sep 16 '13 at 6:49

Since $$\frac{1}{\sqrt{k}} \ge \frac{2}{\sqrt{k}+\sqrt{k+1}} = 2(\sqrt{k+1}-\sqrt{k}) \ge \frac{1}{\sqrt{k+1}}$$

We find. \begin{align} & \sum_{k=1}^{n} \frac{1}{\sqrt{k}} \ge 2\sum_{k=1}^{n}(\sqrt{k+1}-\sqrt{k}) = 2(\sqrt{n+1}-1) \ge 2\sqrt{n} - 2\\ \text{and}\quad & \sum_{k=1}^{n} \frac{1}{\sqrt{k}} = 1 + \sum_{k=1}^{n-1} \frac{1}{\sqrt{k+1}} \le 1 + 2\sum_{k=1}^{n-1}(\sqrt{k+1}-\sqrt{k}) = 2\sqrt{n} - 1 \end{align} As a result, $$2 - \frac{2}{\sqrt{n}} \le S_n \le 2 - \frac{1}{\sqrt{n}} \quad\implies\quad \lim_{n\to\infty} S_n = 2$$

An answer using the Stolz–Cesàro theorem: $$\lim_{n\to\infty} \frac{ \sum_{k=1}^n 1/\sqrt{k} }{\sqrt{n}} = \lim_{n\to\infty} \frac{1/\sqrt{n} }{\sqrt{n} - \sqrt{n-1}} = \lim_{n\to\infty} \frac{\sqrt{n}+\sqrt{n-1} }{\sqrt{n}}=2.$$

\begin{align} {1 \over \sqrt{n}}\,\sum_{k = 1}^{n}{1 \over \sqrt{k\,}} &= {1 \over \sqrt{n}}\,\sum_{k = 1}^{n}{1 \over \sqrt{n\xi_{k}\,}}\,n\Delta\xi = \sum_{k = 1}^{n}{1 \over \sqrt{\xi_{k}\,}}\,\Delta\xi \sim \int_{1/n}^{1}{{\rm d}\xi \over \xi^{1/2}} = \left.\vphantom{\LARGE A}\;2\xi^{1/2}\right\vert_{1/n}^{1} \\[3mm]&= 2\left(1 - {1 \over \sqrt{n\,}}\right) = 2 - {2 \over \sqrt{n\,}} \to \color{#ff0000}{\Large 2} \quad\mbox{when}\quad n \to \infty \end{align}