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.

• What you are after the Euler-Maclaurin summation formula. en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula – Yves Daoust May 3 '17 at 20:00
• indeed the error term can be seen as a sum (or series, depending in the conditions) by the periodicity of the periodic extension of the Bernoulli polynomials. The error term is defined through the Euler-Maclaurin sum formula. – Masacroso May 3 '17 at 20:00
• @Barry Thank you for pointing out that mistake – user3141592 May 3 '17 at 20:35
• @Yves But, applying that formula, where do I get $\zeta(\frac{1}{2})$ from? – user3141592 May 3 '17 at 21:22
• @user3141592: will probably appear as the sum of all terms independent of $n$. – Yves Daoust May 3 '17 at 21:48

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 $\sum\frac1n$.

• Thank you very much! Just to clarify, I do not think that $\frac{1}{k}$ in you second limit of the zeta function is correct, but thanks for the answer, so useful – user3141592 May 4 '17 at 12:56
• Oops, yup, that was a typo. :-) – Simply Beautiful Art May 4 '17 at 13:18
• Just a littlel theoretic question before checking the answer: in the binomial coefficient, as you have a negative fraction ($-1/2$), would you have to apply the definition: $\prod^{2k-1}_{i=1} \frac{1/2-i}{i}$? – user3141592 May 4 '17 at 14:03
• Yes, that is implied most of the time for binomial coefficients with non-natural values. – Simply Beautiful Art May 4 '17 at 18:33

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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