Is this sequence convergent? If so, then how to find the limit?

Question

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.

Answer 1

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 $n \to \infty$.

Answer 2

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

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

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METHODOLOGY $2$: APPLY STOLZ-CESARO

Using the Stolz-Cesaro Theorem, we find that

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