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For each natural number $n$, let $$ x_n \colon= \frac{1}{\sqrt{n}} \left( \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}}+ \cdots + \frac{1}{\sqrt{n}} \right). $$

Then is the sequence $\left( x_n \right)_{n \in \mathbb{N} }$ convergent? And if so, then how to find the limit of this sequence?

I'm just not sure how to proceed.

Saaqib Mahmood
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    Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps. – robjohn Apr 19 '18 at 14:59
  • Are you studying series? In what section did this arise? Are you allowed to use integrals to approximate sums? What have you tried? Please provide context. – robjohn Apr 19 '18 at 15:01

2 Answers2

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METHODOLOGY $1$: USE CREATIVE TELESCOPING

since $\frac12(\sqrt {k-1}+\sqrt k)\le \sqrt k\le \frac12(\sqrt {k+1}+\sqrt k)$, we can use creative telescoping to write

$$\sum_{k=1}^n2\left(\sqrt{k+1}-\sqrt k\right)\le\sum_{k=1}^n\frac{1}{\sqrt k}\le 1+\sum_{k=2}^n 2\left(\sqrt k-\sqrt{k-1}\right)$$

Evaluating the telescoping terms reveals

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

whence dividing by $\sqrt n$ and applying the squeeze theorem yields the result

$$\lim_{n\to \infty }\frac1{\sqrt n}\sum_{k=1}^n \frac1{\sqrt k}=2$$

METHODOLOGY $2$: APPLY STOLZ-CESARO

Using the Stolz-Cesaro Theorem, we find that

$$\begin{align} \lim_{n\to \infty}\frac{\sum_{k=1}^n \frac1{\sqrt k}}{\sqrt n}&=\lim_{n\to \infty} \left(\frac{\frac1{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}}\right)\\\\ &=\lim_{n\to \infty}\left(1+\frac{\sqrt n}{\sqrt{n+1}}\right)\\\\ &=2 \end{align}$$

Mark Viola
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  • On the first line, what happened to the $k=1$ term on the RHS? On the second line, the LHS should be $2(\sqrt{n+1}-1)$, and the RHS should be the same (due to the missing $k=1$ term). –  Apr 19 '18 at 15:06
  • @Bungo The first term is explicitly written as $1$, which is why the summation on the RHS begins at $k=2$. – Mark Viola Apr 19 '18 at 15:09
  • Ah, right, I didn't notice that. Looks good now. (+1) –  Apr 19 '18 at 15:10
  • @Bungo Thank you! Much appreciated. – Mark Viola Apr 19 '18 at 15:10
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This sequence does converge. For every positive integer $n$, since $x \mapsto 1/\sqrt{x}$ is decreasing, we have $$ \int_0^{n} \frac{dx}{\sqrt{x}} > \sum_{k=1}^n \frac{1}{\sqrt{k}} >\int_1^{n+1} \frac{dx}{\sqrt{x}} \\ 2 \sqrt{n} > \sum_{k=1}^n \frac{1}{\sqrt{k}} >2 (\sqrt{n+1}-1) \\ 2 >\frac{1}{\sqrt{n}}\sum_{k=1}^n \frac{1}{\sqrt{k}} >2\left(\sqrt{1+\frac{1}{n}}-\frac{1}{\sqrt{n}}\right) \to 2 $$ as $n \to \infty$.

Chappers
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