How to show that $\sum_{d=1}^n 1/\sqrt{d} - 2 \sqrt{n}$ converges to a constant?

Question

We can approximate $$S_n = \sum_{d=1}^n 1/\sqrt{d}$$ by showing that $$ S_n \le \int_0^n \frac 1 {\sqrt{x}} \mathrm dx = 2 \sqrt{n}, $$ and $$ S_n \ge \int_1^{n+1} \frac 1 {\sqrt{x}} \mathrm dx = 2 \sqrt{n} - O(1). $$

My questions is, is it true that $S_n - 2\sqrt{n}$ converges to a constant?

If this is true, how can we prove it?

Answer 1

We may notice that $$ 2\sqrt{n}-2\sqrt{n-1} = \frac{2}{\sqrt{n}+\sqrt{n-1}}\geq\frac{1}{\sqrt{n}} \tag{1}$$ and the LHS of $(1)$ is a telescopic term. In particular: $$ \left(2\sqrt{n}-2\sqrt{n-1}\right)-\frac{1}{\sqrt{n}} = \frac{1}{\sqrt{n}\left(\sqrt{n}+\sqrt{n-1}\right)^2}\tag{2}$$ and the RHS of $(2)$ is a non-negative and summable (by the p-test) term for $n\geq 1$.
As a consequence, $$ \lim_{n\to +\infty}\left(-2\sqrt{n}+\sum_{k=1}^{n}\frac{1}{\sqrt{k}}\right)=-\sum_{n\geq 1}\frac{1}{\sqrt{n}(\sqrt{n}+\sqrt{n-1})^2}=C<0.\tag{3}$$ Since $\sum_{n\geq 1}\frac{1}{n^s}=\zeta(s)$ for any $s:\text{Re}(s)>1$ and we are regularizing a divergent series, by analytic continuation we have $$ C=\zeta\left(\frac{1}{2}\right) \approx -1.4603545088\tag{4}$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ I'll use the following identity ( from a previous @Vorhang user post ): $$ \sum_{1 \leq\ d\ \leq\ x}{1 \over d^{s}} = {x^{1 - s} \over 1 - s} + \zeta\pars{s} - {\braces{x} \over x^{s}} + s\int_{x}^{\infty}{\braces{u} \over u^{s + 1}} \,\dd u\,,\qquad \Re\pars{s} > 0. $$ $$ \mbox{It leads to}\quad\bbox[#ffd,15px]{ \sum_{d = 1}^{n}{1 \over \root{d}} - 2\root{n} = \zeta\pars{1 \over 2} + {1 \over 2}\int_{n}^{\infty}{\braces{u} \over u^{3/2}}\,\dd u} $$ $$ \mbox{However,}\quad 0 < \verts{{1 \over 2}\int_{n}^{\infty}{\braces{u} \over u^{3/2}}\,\dd u} < {1 \over 2}\int_{n}^{\infty}{\dd u \over u^{3/2}} = {1 \over \root{n}} \implies \lim_{n \to \infty} \bracks{{1 \over 2}\int_{n}^{\infty}{\braces{u} \over u^{3/2}}\,\dd u} = 0 $$

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$$ \mbox{Then,}\quad \bbx{\ds{% \lim_{n \to \infty}\pars{\sum_{d = 1}^{n}{1 \over \root{d}} - 2\root{n}} = \zeta\pars{1 \over 2}}} \approx -1.4604 $$