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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.

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By Cesaro-Stolz

$$\ldots = \lim_{n \to \infty} \frac{\sum_{k=1}^{n+1} \frac{1}{\sqrt{k}}- \sum_{k=1}^{n} \frac{1}{\sqrt{k}} }{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}(\sqrt{n+1} + \sqrt{n})}{n+1 - n} = 2 $$

Also, there is another approach using the squeeze theorem that parallels your argument without appealing to the integral.

Since $\sqrt{k} + \sqrt{k-1} < 2\sqrt{k} < \sqrt{k} + \sqrt{k+1}$ we find the bounds

$$2(\sqrt{k+1} - \sqrt{k})=\frac{2}{\sqrt{k+1} + \sqrt{k}} \leqslant \frac{1}{\sqrt{k}} \leqslant \frac{2}{\sqrt{k} + \sqrt{k-1}} = 2(\sqrt{k} - \sqrt{k-1}).$$

As sums of the LHS and RHS are telescoping, we have

$$2(\sqrt{n+1} - 1) \leqslant \sum_{k=1}^n \frac{1}{\sqrt{k}} \leqslant 2\sqrt{n},$$

and

$$2(\sqrt{1+1/n} - 1/\sqrt{n}) \leqslant \frac{1}{\sqrt{n}}\sum_{k=1}^n \frac{1}{\sqrt{k}} \leqslant 2.$$

Now apply the squeeze theorem.

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  • 3
    $\begingroup$ For those too lazy to look it up: If $a_n$ and $b_n$ are strictly monotone divergent sequences, then $\lim_{n \to \infty} \dfrac{a_{n+1} - a_n}{b_{n+1} - b_n} = \lim_{n \to \infty} \dfrac{a_n}{b_n}$, assuming the former limit exists. Here we have $a_n = 1/\sqrt{1} + \dots + 1/\sqrt{n}$ and $b_n = \sqrt{n}$. $\endgroup$ – MCT Jun 6 '16 at 3:48
  • $\begingroup$ +1 This is precisely the way forward I would have taken with the restriction in the OP. -Mark $\endgroup$ – Mark Viola Jun 6 '16 at 4:13
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From the Euler-Maclaurin Summation Formula, we have

$$\begin{align} \frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{k}}&=\frac{1}{\sqrt{n}}\left(1+\int_1^n \frac{1}{\sqrt{x}}\,dx+\frac12\left(\frac{1}{n^{1/2}}-1\right)+O(1)\right)\\\\ &=2+O\left(\frac{1}{n^{1/2}}\right)\\\\ &\to 2\,\,\text{as}\,\,n\to \infty \end{align}$$

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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:

$$\lim_{n\to\infty}\frac1n\sum_{k=1}^n(k/n)^{-1/2}=\int_0^1x^{-1/2}~\mathrm dx=2x^{1/2}\bigg|_0^1=2$$

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