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It is known (see this link ) that $(1\sqrt{1}) + (1/\sqrt{2})+\cdots + (1/\sqrt{n})\ge \sqrt{n}$ for integers $n\ge 1$.

I took the sequence $x_n=(1\sqrt{1}) + (1/\sqrt{2})+\cdots + (1/\sqrt{n}) -\sqrt{n}$.

First, we can see that this is increasing sequence $$ x_{n+1}-x_n = \frac{1}{\sqrt{n+1}}-\sqrt{n+1}+\sqrt{n} $$ and $\frac{1}{\sqrt{n+1}} \ge \sqrt{n+1}-\sqrt{n}$.

I do not know whether sequence is bounded above or not. Can anyone suggest for it, how to proceed for boundedness and so for convergence.

Maths Rahul
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    you just deleted the same question – Aditya Dwivedi Dec 23 '20 at 06:41
  • If I am not mistaken, $$x_n=\sum_{k=1}^{n}\dfrac{1}{\sqrt{k}}-\sqrt{n}$$ is still divergent. In particular unbounded above. But the modified sequence $$y_n=\sum_{k=1}^{n}\dfrac{1}{\sqrt{k}}-2\sqrt{n}$$ converges to $\zeta\left(\dfrac12\right).$ – Bumblebee Dec 23 '20 at 08:18

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Hint - notice that:

$$ x_{n+1}-x_n = \frac{1}{\sqrt{n+1}}-\sqrt{n+1}+\sqrt{n} = \frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+1}+\sqrt{n}} $$

Andronicus
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$x_n \to \infty$. For this note that $\int_k^{k+1} \frac 1 {\sqrt x} dx <\frac 1 {\sqrt k}$ Sum this from $k=1$ to $n-1$ to see that $x_n >\frac 1 {\sqrt n} +\sqrt n-2$. [Note that $\int_1^{n} \frac 1 {\sqrt x} dx=2\sqrt n -2$]

Kavi Rama Murthy
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