<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] At least, for "large" values on $n$, we can show it since
$$\sum_{i=n+2}^{+\infty}\dfrac{1}{i^2}=\psi ^{(1)}(n+2)$$
So, using the asymptotics of the digamma function and Taylor expansions, we have
$$\sum_{i=n+2}^{+\infty}\frac{1}{i^2}-\frac{2n+5}{2(n+2)^2}=\frac{1}{6 n^3}+O\left(\frac{1}{n^4}\right)$$
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