Closed form for $\sum_k \frac{1}{\sqrt{k}}$

Question

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.

Answer 1

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 $\sum\frac1n$.

Answer 2

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