<s>[INST] How prove this $\sum_{i=n+2}^{+\infty}\frac{1}{i^2}>\frac{2n+5}{2(n+2)^2}$ let $n$ be postive integer,show that
$$\sum_{i=n+2}^{+\infty}\dfrac{1}{i^2}>\dfrac{2n+5}{2(n+2)^2}\tag{1}$$
I know
$$\sum_{i=n+2}^{+\infty}\dfrac{1}{i^2}>\int_{n+2}^{+\infty}\dfrac{1}{x^2}dx=\dfrac{1}{n+2}$$
But $$\dfrac{1}{n+2}-\dfrac{2n+5}{2(n+2)^2}=-\dfrac{1}{2(n+2)^2}<0$$
then this integral method can't solve (1),so How  to prove it?Thanks
 [/INST] There is a rather elegant solution, which uses the convexity of the function. Since $f(x) = \frac{1}{x^2}$ is convex, let us consider an area element under the curve versus the area of the trapezoid formed by joining $(n+2,f(n+2))$ and $(n+3,f(n+3)$ by a straight line
$$dA < \frac{1}{2}\left(\frac{1}{(n+2)^2} + \frac{1}{(n+3)^2}\right)$$
$$\implies \int_{n+2}^\infty \frac{1}{x^2}dx < \frac{1}{2(n+2)^2} + \sum_{n+3}^\infty \frac{1}{i^2}$$
$$\implies \frac{1}{n+2} + \frac{1}{2(n+2)^2} < \sum_{n+2}^\infty\frac{1}{i^2}$$
EDIT
Added an illustration of the above just to clarify how I got the inequality

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