Find formula for $\frac{1}{\sqrt 1}+ \frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt n}$ I have the series:
$$\frac{1}{\sqrt 1}+ \frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt n}$$
I find hard to generalize into one formula, any explanation would be helpful.
 A: Analogous to Euler-Mascheroni Constant we have:
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}=
2\sqrt{n}-1.4603545088\ldots+\frac{1}{2\sqrt{n}}-\frac{1}{24\sqrt{n^{3}}}+O\left( \frac{1}{\sqrt{n^{7}}} \right)$$
where $\displaystyle
\lim_{n\to \infty} \left( 2\sqrt{n}-1-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}-
\ldots-\frac{1}{\sqrt{n}} \right) = 1.4603545088\ldots$
A: A decent aproximation is:
$$ \frac{1}{\sqrt1} + \frac{1}{\sqrt2}+⋯+\frac{1}{\sqrt n} \simeq 2\sqrt n $$
Let's calculate the limit of the quotient using 's rule:
$$\lim_{n \rightarrow \infty} \frac{\sum \limits_{k=1}^{n} \frac{1}{\sqrt k}}{2\sqrt n} = \lim_{n \rightarrow \infty}\frac{\frac{1}{\sqrt n}}{2(\sqrt n - \sqrt {n-1})} = \frac{1}{2} \lim_{n \rightarrow \infty} \frac{\sqrt{n} + \sqrt{n - 1}}{\sqrt n} = 1$$
This means that the relative error between the two functions is small for big values of $n$, and therefore they are equivalent in a limit when $n \rightarrow \infty$
A: Let the above sum = $S$ then $S \approx \displaystyle \int_1^{n}\dfrac{1}{\sqrt{n}}dn$  = $2(\sqrt{n} - 1)$
Remember that this not a general term/formula, this is just an approximation.
A: One might note the following approximations:
$$\sum_{k=1}^n\frac1{\sqrt k}\approx \zeta(1/2)+\sqrt n+\sqrt{n+1}$$
and
$$\sum_{k=1}^n\frac1{\sqrt k}\approx \zeta(1/2)+2\sqrt{n+\frac12}$$
where $\zeta(1/2)\approx-1.4603545088$
Here is a table for a few values:
$$\begin{array}{c|c|c|c}n&\sum_{k=1}^n\frac1{\sqrt k}&\sqrt n+\sqrt{n+1}&2\sqrt{n+\frac12}\\\hline3&2.2844570504&2.2716962988&2.2813028780\\9&4.7047701338&4.7019231514&4.7040594942\\27&9.0278784160&9.0273005360&9.0277339729\\81&16.5951438914&16.5950306293&16.5951155765\end{array}$$
Better approximations may be obtained with the Euler-Maclaurin summation formula.
