Michelle Mendoza

2022-07-05

Show that for every integer $1+\frac{1}{4}+\frac{1}{9}+···+\frac{1}{{n}^{2}}\le 2-\frac{1}{n}$

Kiana Cantu

Expert

Hint: Note that
$\frac{1}{{2}^{2}}+\frac{1}{{3}^{2}}+\cdots +\frac{1}{{n}^{2}}<\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots +\frac{1}{\left(n-1\right)\cdot n}.$
The expression on the right turns out to be a telescoping sum. For $\frac{1}{\left(i-1\right)\cdot i}=\frac{1}{i-1}-\frac{1}{i}$
One could also do an induction version of the above proof, using the fact that $\frac{1}{\left(n+1{\right)}^{2}}<\frac{1}{n}-\frac{1}{n+1}$

ban1ka1u

Expert

Consider a Riemann sum approximation for
${\int }_{1}^{n}\frac{1}{{x}^{2}}dx$
by means of the partition $\left\{1,2,3,\dots ,n\right\}$ and a right endpoint approximation. Since $\frac{1}{{x}^{2}}$ is decreasing, the approximation will give an underestimate for the integral.
We obtain
$\frac{1}{4}+\frac{1}{9}+\cdots +\frac{1}{{n}^{2}}\le {\int }_{1}^{n}\frac{1}{{x}^{2}}dx={\left[-\frac{1}{x}\right]}_{1}^{n}=1-\frac{1}{n}$
The result in the problem follows immediately.

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