limit( Sigma( sqrt(1/k),k goes from 1 to n), n goes to infinity)
What are you trying to prove? I see no propositions in your post. Are you trying to show divergence? We have
$\displaystyle \sum_{k=1}^\infty\sqrt{\frac1k} = \sum_{k=1}^\infty\frac1{\sqrt{k}} = \sum_{k=1}^\infty k^{-1/2}$
This is a p-series, with $\displaystyle p = \frac12$, so you should know that it diverges. If you want to prove it without relying on the properties of p-series, one way is to use the Integral Test. Show that $\displaystyle \int_1^\infty x^{-1/2}\,dx$ diverges (and note that $\displaystyle x^{-1/2}$ is positive, decreasing, and continuous on $\displaystyle [1,\,\infty)$).
Hello
As $\displaystyle x\mapsto \frac{1}{\sqrt{x}}$ is a decreasing function, you can use $\displaystyle \int_{k}^{k+1}\frac{\mathrm{d}t}{\sqrt{t}} \leq \int_{k}^{k+1}\frac{\mathrm{d}t}{\sqrt{k}}\, \leq \int_{k-1}^{k}\frac{\mathrm{d}t}{\sqrt{t}} \Longleftrightarrow
\int_{k}^{k+1}\frac{\mathrm{d}t}{\sqrt{t}} \leq \frac{1}{\sqrt{k}} \leq \int_{k-1}^{k}\frac{\mathrm{d}t}{\sqrt{t}} $.
Adding these inequalities for $\displaystyle k\in[|2,\,n|]$ gives us :
$\displaystyle \int_{2}^{3}\frac{\mathrm{d}t}{\sqrt{t}}+ \int_{3}^{4}\frac{\mathrm{d}t}{\sqrt{t}}+\ldots+ \int_{n}^{n+1}\frac{\mathrm{d}t}{\sqrt{t}}
\leq \sum_{k=2}^n\frac{1}{\sqrt{k}}
\leq
\int_{1}^{2}\frac{\mathrm{d}t}{\sqrt{t}}+ \int_{2}^{3}\frac{\mathrm{d}t}{\sqrt{t}}+\ldots+ \int_{n-1}^{n}\frac{\mathrm{d}t}{\sqrt{t}}$
Adding 1 yields
$\displaystyle 1+\int_{2}^{n+1}\frac{\mathrm{d}t}{\sqrt{t}}
\leq \sum_{k=1}^n\frac{1}{\sqrt{k}}
\leq
1+\int_{1}^{n}\frac{\mathrm{d}t}{\sqrt{t}}$
Can you take it from here ?
$\displaystyle \lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum\limits_{k\ ,=\,1}^{n}{\frac{1}{\sqrt{k}}}=\lim_{n\to\infty}\f rac{1}{n}\sum\limits_{k\,=\,1}^{n}{\sqrt{\frac{n}{ k}}}.$
For each $\displaystyle n,$ it's a Riemann sum for $\displaystyle \int_{0}^{1}{\frac{1}{\sqrt{x}}\,dx}$ and the conclusion follows.