Vtyjkyuk

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?

HINT

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

The intuition here is that $$ sum_nge x1/n^2 sim int_x^infty frac1t^2dt = frac1x.$$ 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 $c > 1.$

Practically we can say $xin[2,infty)$ then from $n>n-1$ we have
$$sum_ngeq xdfrac1n^2<sum_ngeq xdfrac1n(n-1)=sum_ngeq xleft(dfrac1n-1-dfrac1nright)<dfrac1x-1leqdfrac2x$$