<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] By using $\frac{1}{q^2} = \int_0^\infty \mathrm{e}^{-x q} x \mathrm{d} x$ for $q > 0$, we have (the summation and the integral are interchangeable by Fubini/Tonelli theorems)
\begin{align}
\sum_{i=n+2}^\infty \frac{1}{i^2} &= \sum_{i=n+2}^\infty \int_0^\infty \mathrm{e}^{-x i} x \mathrm{d} x\\
&=  \int_0^\infty \sum_{i=n+2}^\infty \mathrm{e}^{-x i} x \mathrm{d} x\\
&= \int_0^\infty \mathrm{e}^{-x(n+2)} \frac{x}{ 1 - \mathrm{e}^{-x}} \mathrm{d} x\\
&> \int_0^\infty \mathrm{e}^{-x(n+2)} \left(1 + \frac{x}{2}\right) \mathrm{d} x\\
&= \frac{2n+5}{2(n+2)^2}
\end{align}
where we have used
$$\sum_{i=n+2}^\infty \mathrm{e}^{-x i} = \mathrm{e}^{-x(n+2)} \frac{1}{ 1 - \mathrm{e}^{-x}}$$
and (see the remark later)
$$\frac{x}{ 1 - \mathrm{e}^{-x}} > 1 + \frac{x}{2}, \ \forall x > 0. \tag{1}$$
Remark 1: To prove (1), it suffices to prove that $\mathrm{e}^{-x} > \frac{2-x}{2+x}$  or $-x > \ln \frac{2-x}{2+x}$ for $x \in (0, 2)$.
Let $f(x) = -x - \ln \frac{2-x}{2+x}$. We have $f'(x) = \frac{x^2}{(2-x)(2+x)} > 0$ for $x \in (0, 2)$. Also, $f(0) = 0$. Thus, we have $f(x) > 0$ for $x\in (0, 2)$.
Remark 2: $1 + \frac{x}{2}$ is first two terms of the Taylor expansion of $\frac{x}{ 1 - \mathrm{e}^{-x}}$ around $x = 0$.
</s>