Existence of a constant in a convergent series.

Question

Prove that $\exists$ a constant $c > 0$ such that for all $x \in [1, \infty)$, $$ \sum_{n \geq x} 1/ n^2 \leq c/x $$.

What I can understand that $\sum 1/n^2$ is convergent and it converges to $\pi^2 / 6$. So it must be bounded. Here tail of the series is convergent.

Whether I can take a $c > 0$ such that $Mx < c$ where $M$ is the bound of the series?

Answer 1

HINT

You cannot do that since $Mx >c$ for sufficiently large $x.$ Because $1/x\to 0,$ this requires a better estimate than simply the fact that the sum converges.

The intuition here is that $$ \sum_{n\ge x}1/n^2 \sim \int_x^\infty \frac{1}{t^2}dt = \frac{1}{x}.$$ Now you just need to analyze the error in this approximation to get the inequality you need. Note that the sum is larger than $1/x,$ so you'll need to choose a $c > 1.$

Answer 2

Practically we can say $x\in[2,\infty)$ then from $n>n-1$ we have $$\sum_{n\geq x}\dfrac{1}{n^2}<\sum_{n\geq x}\dfrac{1}{n(n-1)}=\sum_{n\geq x}\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right)<\dfrac{1}{x-1}\leq\dfrac{2}{x}$$