Show that for every integer $n ≥ 1$, $1 + \frac{1}{4} +\frac{1}{9} + · · · + \frac{1}{n^2} ≤ 2 − \frac{1}{n}$

Question

I can just think of trying to prove $\frac{1}{4} +\frac{1}{9} + · · · + \frac{1}{n^2} ≤ \frac{1}{n}$, but remains stuck.

Answer 1

Hint: Note that $$\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{n^2} \lt \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots+\frac{1}{(n-1)\cdot n}.$$ The expression on the right turns out to be a telescoping sum. For $\frac{1}{(i-1)\cdot i}=\frac{1}{i-1}-\frac{1}{i}$.

One could also do an induction version of the above proof, using the fact that $\frac{1}{(n+1)^2}\lt \frac{1}{n}-\frac{1}{n+1}$.

Answer 2

Consider a Riemann sum approximation for $$\int_1^n \frac{1}{x^2} dx$$ by means of the partition $\{1, 2, 3,\dots, n\}$ and a right endpoint approximation. Since $\frac1{x^2}$ is decreasing, the approximation will give an underestimate for the integral.

We obtain $$\frac14+\frac19+\cdots+\frac1{n^2}\le \int_1^n \frac{1}{x^2}dx=\left[-\frac1x\right]_1^n=1-\frac1n$$

The result in the problem follows immediately.