An interesting inequality $\sum_{k=1}^n \frac{1}{n+k}<\frac{\sqrt{2}}{2}, \ n\ge1$

Question

Here is one of the beautiful inequalities from Elementary inequalities by Mitrovic $$\sum_{k=1}^n \frac{1}{n+k}<\frac{\sqrt{2}}{2},$$ which is easy to prove by calculus using that $\lim_{n\to\infty}\sum_{k=1}^{n} \frac{1}{n+k}=\log(2)$.

Now, the question is How would you prove it without calculus?

Answer 1

Cauchy-Schwarz plus creative telescoping and a bit of luck:

$$\sum_{k=1}^{n}\frac{1}{n+k}<\sum_{k=1}^{n}\frac{1}{\sqrt{n+k-1}\sqrt{n+k}}\stackrel{CS}{\leq}\sqrt{n\sum_{k=1}^{n}\left(\frac{1}{n+k-1}-\frac{1}{n+k}\right)}=\color{red}{\frac{1}{\sqrt{2}}}.$$

Answer 2

(Alternatively) By using Root-Mean Square-Arithmetic Mean we get that

$$\sum_{k=1}^{n}\frac{1}{n+k}<\sqrt{n\sum_{k=1}^{n}\frac{1}{(n+k)^2}}<\sqrt{n\sum_{k=1}^{n}\frac{1}{(n+k)(n+k-1)}}=\frac{1}{\sqrt{2}}.$$

Answer 3

Using Bernoulli's Inequality, which can be proven by induction, we get that for $x\ge-n$, $$ \left(1+\frac{x}{n}\right)^n\ge1+x $$ Taking the limit as $n\to\infty$, we get that for all $x\in\mathbb{R}$, $$ e^x\ge1+x $$ which implies that for $x\in(0,1)$, $$ x\le-\log(1-x) $$ Therefore, $$ \begin{align} \sum_{k=1}^n\frac1{n+k} &\le\sum_{k=1}^n\log\left(\frac{n+k}{n+k-1}\right)\\ &=\log(2)\\ &\lt\frac{\sqrt2}2 \end{align} $$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = 1}^{n}{1 \over n + k} & = \sum_{k = n + 1}^{2n}{1 \over k} = \sum_{k = 1}^{2n}{1 \over k} - \sum_{k = 1}^{n}{1 \over k} = H_{2n} - H_{n}\,, \qquad\pars{~H_{z}: Harmonic\ Number~} \end{align}

Since $\ds{\pars{~\gamma:\ Euler\!-\!Mascheroni\ Constant~}}$

$$ \ln\pars{m + {1 \over 2}} + \gamma + {1 \over 24\pars{m + 1}^{2}} < H_{m} < \ln\pars{m + {1 \over 2}} + \gamma + {1 \over 24m^{2}} $$ \begin{align} &\mbox{I'll have}\quad\left\{\begin{array}{rcl} \ds{H_{2n}} & \ds{<} & \ds{\ln\pars{2n + {1 \over 2}} + \gamma + {1 \over 96n^{2}}} \\[2mm] \ds{-H_{n}} & \ds{<} & \ds{-\ln\pars{n + {1 \over 2}} - \gamma - {1 \over 24\pars{n + 1}^{2}}} \end{array}\right. \\[5mm] & \implies \bbx{\left.\sum_{k = 1}^{n}{1 \over n + k}\right\vert_{\ n\ \geq\ 1} < \ln\pars{2 - {1 \over 2n + 1}} - {\pars{n + 1/3}\pars{n - 1} \over 32n^{2}\pars{n + 1}^{2}} <\ \ln\pars{2} < \color{#f00}{{\root{2} \over 2}}} \end{align}

Indeed, $\ds{\ln\pars{2} \approx 0.6931}$ is a "better bound" that $\ds{{\root{2} \over 2} \approx 0.7071}$ !!!.