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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

math110
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3 Answers3

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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

enter image description here

Dhanvi Sreenivasan
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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 $x = 0$.

River Li
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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 $$\sum_{i=n+2}^{+\infty}\frac{1}{i^2}-\frac{2n+5}{2(n+2)^2}=\frac{1}{6 n^3}+O\left(\frac{1}{n^4}\right)$$

Claude Leibovici
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