limit $\lim\limits_{n\to\infty}\left(\sum\limits_{i=1}^{n}\frac{1}{\sqrt{i}} - 2\sqrt{n}\right)$ Calculate below limit
$$\lim_{n\to\infty}\left(\sum_{i=1}^{n}\frac{1}{\sqrt{i}} - 2\sqrt{n}\right)$$

The following is an elementary consideration, which shows how to compute the limit in terms of an infinite series. It's evaluation requires usage of Euler's summation formula, already covered by Dane.
Consider the following transformation
$$
   \sum_{k=1}^n \frac{1}{\sqrt{k}} = \sum_{k=1}^n \left(\frac{1}{\sqrt{k}} - \frac{2}{\sqrt{k} + \sqrt{k+1}} \right) + \sum_{k=1}^n \frac{2}{\sqrt{k} + \sqrt{k+1}}
$$
Then use $\sqrt{k+1}-\sqrt{k} = \frac{\left(\sqrt{k+1}-\sqrt{k}\right)\left(\sqrt{k+1}+\sqrt{k}\right)}{\sqrt{k+1}+\sqrt{k}} = \frac{(k+1)-k}{\sqrt{k+1}+\sqrt{k}} = \frac{1}{\sqrt{k+1}+\sqrt{k}}$:
$$
  \sum_{k=1}^n \frac{1}{\sqrt{k}} = \sum_{k=1}^n \frac{1}{\sqrt{k} \left( \sqrt{k} + \sqrt{k+1} \right)^2} + 2 \sum_{k=1}^n \left( \sqrt{k+1}-\sqrt{k} \right)
$$ 
The latter sum telescopes:
$$
   \sum_{k=1}^n \left( \sqrt{k+1}-\sqrt{k} \right) = \left( \sqrt{2}-\sqrt{1} \right) + \left( \sqrt{3}-\sqrt{2} \right) + \cdots + \left( \sqrt{n+1}-\sqrt{n} \right) = \sqrt{n+1}-1 
$$
From here:
$$ \begin{eqnarray}
   \left(\sum_{k=1}^n \frac{1}{\sqrt{k}} \right)- 2 \sqrt{n} &=& 
       \sum_{k=1}^n \frac{1}{\sqrt{k} \left( \sqrt{k} + \sqrt{k+1} \right)^2} + 2 \left( \sqrt{n+1}-\sqrt{n}-1\right) \\
    &=& \sum_{k=1}^n \frac{1}{\sqrt{k} \left( \sqrt{k} + \sqrt{k+1} \right)^2} + 2 \left( \frac{1}{\sqrt{n+1}+\sqrt{n}}-1\right) 
\end{eqnarray}
$$
In the limit:
$$
  \lim_{n\to \infty} \left(\sum_{k=1}^n \frac{1}{\sqrt{k}} \right)- 2 \sqrt{n} = -2 + 
      \sum_{k=1}^\infty \frac{1}{\sqrt{k} \left( \sqrt{k} + \sqrt{k+1} \right)^2}
$$
