Prove the following sequence is Cauchy

Question

I was proving the following sequence converges using Cauchy’s theorem as required: $$ x_n=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}-2\sqrt{n}. $$ I can actually prove that it converges through proving it decreases and is bounded, but the question requires to prove it by Cauchy and I did not figure it out. Can anybody give me a hand?

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I did the following job: in order to find a proper $N\in\mathbb{N}_+$ for a fixed $\epsilon>0$, calculate (supposing $m>n$) \begin{align*} |x_m-x_n|={}&\left|\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+\ldots+\frac{1}{\sqrt{m}}-2\sqrt{m}+2\sqrt{n}\right|\\ \leq{}&\left|\frac{m-n}{\sqrt{n}}-2\frac{m-n}{\sqrt{m}+\sqrt{n}}\right|, \end{align*} and I got stuck… I can’t let $m$ disappear and think only about $n$, which means I’m not able to find the $N$ as required.. Did I go too far on zooming the inequality?

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So I followed the hint by @Gribouillis and get: \begin{align*} x_n-x_{n+1}={}&-2\sqrt{n}+2\sqrt{n+1}-\frac{1}{\sqrt{n+1}}\\ ={}&\frac{2}{\sqrt{n}+\sqrt{n+1}}-\frac{1}{\sqrt{n+1}} \end{align*} Which leads to \begin{align*} 0\le x_n-x_{n+1}\le\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}, \end{align*} And this immediately leads to $$ |x_m-x_n|\le\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{m}}<\frac{1}{\sqrt{n}},\qquad (\text{supposing}\ m>n) $$ So I can simply let $N=\left[\frac{1}{\epsilon^2}\right]+1$, thus $|x_n-x_m|<\epsilon$. Thanks again to everyone commented below.

Answer 1

Hint: prove that \begin{equation} 0 \le x_n - x_{n+1}\le \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} \end{equation}

Answer 2

Noting that \begin{eqnarray} x_n&=&1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}-2\sqrt{n}\\ &=&\sum_{k=1}^n\int_{k-1}^k\frac{1}{\sqrt{k}}dx-\int_1^n\frac{1}{\sqrt{x}}dx+2\\ &=&\sum_{k=1}^n\int_{k-1}^k(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{x}})dx+2 \end{eqnarray} one has, for $p>0$, \begin{eqnarray} |x_{n+p}-x_n|&=&\bigg|\sum_{k=n+1}^{n+p}\int_{k-1}^k(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{x}})dx\bigg|\\ &\le&\sum_{k=n+1}^{n+p}\int_{k-1}^k(\frac{1}{\sqrt{k-1}}-\frac{1}{\sqrt{k}})dx\\ &=&\sum_{k=n+1}^{n+p}(\frac{1}{\sqrt{k-1}}-\frac{1}{\sqrt{k}})\\ &=&\frac{1}{\sqrt n}-\frac1{\sqrt{n+p}}\\ &=&\frac{1}{\sqrt n}. \end{eqnarray} So for $\epsilon>0$, letting $N=\left[\frac{1}{\epsilon^2}\right]+1$, one has $$|x_{n+p}-x_n|<\epsilon.$$