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A function $f$ is said to have a lower bound $c$ if $c \le f(x)$ for all $x$ in its domain. The definition is similar for sequences.

Now Assume that the sequence $a_n=\sum_{k=1}^{n} \frac{1}{\sqrt k} - 2\sqrt n$ is given.
How can we prove that this sequence has lower bound?

Note : It's much easy to show that the sequence has upper bound. But, my problem is that it seems i have nothing to work on in order to prove the existance of lower bound.

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  • $\begingroup$ By the reflection formula for the $\zeta$ function, $$\lim_{n\to +\infty}\left(-2\sqrt{n}+\sum_{k=1}^{n}\frac{1}{\sqrt{k}}\right) = \zeta\left(\frac{1}{2}\right) = -1.4603545088\ldots $$ $\endgroup$ – Jack D'Aurizio Oct 17 '16 at 17:52
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Hint: Observe \begin{align} \sum^n_{k=1} \frac{1}{\sqrt{k}} \geq \int^{n+1}_1 \frac{1}{\sqrt{x}}\ dx = 2\sqrt{n+1}-2. \end{align}

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  • $\begingroup$ Great !!! How did you think about integration ? $\endgroup$ – Arman Malekzadeh Oct 17 '16 at 17:18
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Variation without integration. Going from $n$ to $n+1$ adds $$\begin{eqnarray}\frac1{\sqrt{n+1}}-2\left(\sqrt{n+1}-\sqrt{n}\right) &=& \frac1{\sqrt{n+1}} -\frac2{\sqrt{n}+\sqrt{n+1}}\\ &>& \frac1{\sqrt{n+1}} -\frac1{\sqrt{n}} \end{eqnarray}.$$ Starting at $a_1=-1$ this results in a telescoping sum for $n > 1$ $$\sum_{k=1}^n\frac1{\sqrt{k}} - 2\sqrt{n}>-2+\frac1{\sqrt{n}}.$$

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