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