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Show directly (from the definition) that if

$$a_n=1+\frac{1}{\sqrt{2}} +\frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}}\;,$$

then $(a_n)$ is not a Cauchy sequence.

Attempt

Using $a_n$ is not Cauchy if $$\exists\, \varepsilon > 0 : \forall N\, \exists\, n, m > N | a_{n} - a_{m}| \geq \varepsilon$$

taking $n>m$.

$$\begin{align}|a_n - b_m| = \left|\sum_{k=1}^{n} \frac{1}{\sqrt{k}} - \sum_{l=1}^{m} \frac{1}{\sqrt{l}}\right| &= \left|\sum_{k=m+1}^{n} \frac{1}{\sqrt{k}}\right| \ge \left|\sum_{k=m+1}^{n} \frac{1}{n}\right| = \frac{n-m}{n}\\ \end{align} $$

Letting $n = 2m$

$$\frac{n-m}{n} = \frac{2m-m}{2m} = \frac{m}{2m} = \frac{1}{2}$$

Thus

$$\exists\, \varepsilon = \frac{1}{2} : \forall N\, \exists\, n=2m > m > N : \left|\sum_{k=m+1}^{n} \frac{1}{\sqrt{k}}\right| \geq \frac{1}{2}$$

so $a_n$ is not Cauchy.

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    $\begingroup$ Bravo!!!!!!!!!!! $\endgroup$ – mathcounterexamples.net Oct 19 '15 at 17:06
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    $\begingroup$ You haven't actually specifically chosen $n,m>N$, but the proof is okay otherwise. You should say "Letting $m=N+1,n=2m$,..." $\endgroup$ – Thomas Andrews Oct 19 '15 at 17:16
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Looks like a very well executed proof. There's only one typo, since $|a_n-b_m|$ should actually be $|a_n - a_m|$.

Also, you can just drop the absolute values at $$\sum_{k=m+1}^n\frac{1}{\sqrt k}$$

since the value is obviously positive.

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Another proof:

If $a_n =\sum_{k=1}^n \frac1{\sqrt{k}} $, $a_{n^2}-a_n =\sum_{k=n+1}^{n^2} \frac1{\sqrt{k}} >\sum_{k=n+1}^{n^2} \frac1{\sqrt{n^2}} =(n^2-n)\frac1{n} =n-1 $.

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I will not answer this question but I provide another solution to it, $\forall n \geq 1\leftrightarrow \frac{1}{n} \leq 1 \leftrightarrow \frac{1}{\sqrt{n}} \geq \frac{1}{n} \leftrightarrow \sum\limits_{n=1}^{+\infty} \frac{1}{\sqrt{n}} \geq \sum \limits_{n=1}^{+\infty} \frac{1}{n}$ knowing that $\sum \limits_{n=1}^{+\infty} \frac{1}{n}$ diverges, all will the sum presented in the OP so it is not cauchy !

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