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

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

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

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

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

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, $$ \begin{align} \sum_{k=1}^n\frac1{n+k} &\le\sum_{k=1}^n\log\left(\frac{n+k}{n+k-1}\right)\\ &=\log(2)\\ &\lt\frac{\sqrt2}2 \end{align} $$