# Prove $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdot\cdot\cdot+\frac{1}{x^2}<\frac{x-1}{x}$

Please help me for proving this inequality $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdot\cdot\cdot+\frac{1}{x^2}<\frac{x-1}{x}$

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Hint: induction – Thomas Andrews Sep 9 '12 at 19:17
Hint: $\frac1{3^2} < \frac1{2\cdot3} = \frac12-\frac13$. Try to do the same thing for all summands. – Martin Sleziak Sep 9 '12 at 19:18
related to this thread since $\frac {\pi^2}6-1<1$. – Raymond Manzoni Sep 9 '12 at 19:49

$$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdot\cdot\cdot+\frac{1}{x^2}<\frac{x-1}{x}$$

Employ the inequality: $$\frac{1}{a^2}=\frac{1}{a}\cdot\frac{1}{a}<\frac{1}{a}\cdot\frac{1}{a-1}=\frac{1}{a-1}-\frac{1}{a}$$.

In $$\displaylines{ \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + \cdots + \frac{1}{{{x^2}}} < \cr \left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + \cdots + \left( {\frac{1}{{x - 1}} - \frac{1}{x}} \right) = \cr = 1 - \frac{1}{x} = \frac{{x - 1}}{x} \cr}$$

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A more complicated approach is to prove the result by induction. Let $$S(n)=\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{n^2}.$$ We want to show that $S(n)\lt 1-\dfrac{1}{n}$ for every integer $n \ge 2$.

The result is clearly true when $n=2$. We show that for any $k\ge 2$, if the result is true when $n=k$, it is true when $n=k+1$. We have
$$S(k+1)=S(k)+\frac{1}{(k+1)^2}.$$ By the induction assumption, $S(k)\lt 1-\dfrac{1}{k}$. It follows that $$S(k+1)\lt 1-\frac{1}{k}+\frac{1}{(k+1)^2}.$$ But $$\frac{1}{k}-\frac{1}{(k+1)^2}=\frac{k^2+k+1}{k(k+1)^2}\gt \frac{k^2+k}{k(k+1)^2}=\frac{1}{k+1}.$$ It follows that $S(k+1)\lt 1-\dfrac{1}{k+1}$.

Remark: If we try to prove the weaker result $S(n)\lt 1$ by induction, we get into trouble. For from the induction assumption $S(k)\lt 1$, we cannot conclude in any direct way that $S(k+1)\lt 1$. So with the above problem, we have the seemingly paradoxical fact that in an induction proof, a strong inequality can be easier to prove than a weaker inequality.

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