How is this true?
$\frac{\sqrt{k}}{k^2}+\frac{\sqrt{k+1}}{(k+1)^2}+\cdots+\frac{\sqrt{n}}{n^2} \leqslant \int_{k-\frac{1}{2}}^{\infty} \frac{\sqrt{x}}{x^2} d x$
where $k,n$ are natural numbers. I can only think of the upper bound as $\int_{k-1}^{\infty} \frac{\sqrt{x}}{x^2} d x$. Where does this 1/2 come from?
Btw, this is used in proving $\frac{\sqrt{k}}{k^2}+\frac{\sqrt{k+1}}{(k+1)^2}+\cdots+\frac{\sqrt{n}}{n^2} \leqslant 4(\sqrt{k}-\sqrt{k-1})$
Assume a function $f(x)$ is convex on $[-1/2,1/2].$ Then $$\int\limits_{-1/2}^{1/2}f(x)\,dx\ge f(0)\quad (*) $$ Indeed $$\int\limits_{-1/2}^{1/2}f(x)\,dx=\int\limits_{0}^{1/2}[f(x)+f(-x)]\,dx$$ By convexity $$f(x)+f(-x)\ge 2f(0)$$ This gives $(*).$
The function $x^{-3/2}$ for $x>0$ is convex as it's second derivative is positive. Therefore applying the first part to $f(x)=(x+k)^{-3/2}$ yields $$\int\limits_{k-1/2}^{k+1/2}x^{-3/2}\,dx \ge k ^{-3/2}$$