Closed form for $\sum_k \frac{1}{\sqrt{k}}$ I would like to know if there's any equivalence to:
$$\sum_{k=1}^n \frac{1}{k} = \log n + \gamma + \frac{1}{2n} - \sum_{k=1}^\infty \frac{B_{2k}}{2kn^{2k}}$$
But to define $E(n)$ in:
$$\sum_{k=1}^n \frac{1}{k^{1/2}} =2  \sqrt{n} + \zeta(\frac{1}{2}) + \frac{1}{2\sqrt{n}} + E(n)$$
(For $n$ an integer $n>1$)
I would like to express the error term as a sum/series rather than an integral. I did not find anything on the Internet but error terms using big-O notation. The only exact formula I found is: 
$$\sum_{k=1}^n \frac{1}{k^{1/2}} = \zeta(\frac{1}{2}) - \zeta (\frac{1}{2}, n+1)$$
I do not know hoy to get from there to $E(x)$. Any help?
Thank you.
 A: According to the Euler-Maclaurin summation formula, we have
$$\sum_{k=1}^n\frac1{\sqrt k}=2\sqrt n+\zeta(1/2)+\frac1{2\sqrt n}+\sum_{k=1}^\infty\frac{B_{2k}}{(2k)!}\binom{-1/2}{2k-1}n^{\frac12-2k}$$
Notice that all the constant terms get turned into $\zeta(1/2)$, which follows since:
$$\zeta(1/2)=\lim_{n\to\infty}\sum_{k=1}^n\frac1{\sqrt k}-\int_0^n\frac1{\sqrt x}\ dx\\\zeta(1/2)=\lim_{n\to\infty}\sum_{k=1}^n\frac1{\sqrt k}-\int_0^n\frac1{\sqrt x}\ dx-\underbrace{\frac1{2\sqrt n}-\sum_{k=1}^\infty\frac{B_{2k}}{(2k)!}\binom{-1/2}{2k-1}n^{\frac12-2k}}_{\to0}$$
Notice the similarity to your other result for $\sum\frac1n$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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Another one is given by
  a well known identity:

\begin{align}
\sum_{k = 1}^{n}{1 \over k^{1/2}} & =
2\root{n} + \zeta\pars{1 \over 2} +
{1 \over 2}\int_{n}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x
\end{align}

Note that
  $\ds{0 <
{1 \over 2}\int_{n}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x <
{1 \over 2}\int_{n}^{\infty}{\dd x \over x^{3/2}} = {1 \over \root{n}}}$ such that

$$
2\root{n} +  \zeta\pars{1 \over 2} <
\sum_{k = 1}^{n}{1 \over k^{1/2}} <
2\root{n} + {1 \over \root{n}} + \zeta\pars{1 \over 2}
$$
and
$$
\sum_{k = 1}^{n}{1 \over k^{1/2}} \approx
2\root{n} + {1 \over 2\root{n}} + \zeta\pars{1 \over 2}\quad
\mbox{with an}\ absolute\ error\ < {1 \over 2\root{n}}
$$
