An interesting inequality $\sum_{k=1}^n \frac{1}{n+k}<\frac{\sqrt{2}}{2}, \ n\ge1$ 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?
 A: (Alternatively) By using Root-Mean Square-Arithmetic Mean we get that 
$$\sum_{k=1}^{n}\frac{1}{n+k}<\sqrt{n\sum_{k=1}^{n}\frac{1}{(n+k)^2}}<\sqrt{n\sum_{k=1}^{n}\frac{1}{(n+k)(n+k-1)}}=\frac{1}{\sqrt{2}}.$$
A: 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,
$$
\begin{align}
\sum_{k=1}^n\frac1{n+k}
&\le\sum_{k=1}^n\log\left(\frac{n+k}{n+k-1}\right)\\
&=\log(2)\\
&\lt\frac{\sqrt2}2
\end{align}
$$
A: Cauchy-Schwarz plus creative telescoping and a bit of luck:

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

A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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\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 $\ds{{\root{2} \over 2} \approx 0.7071}$ !!!.

