proving a hard inequality

Question

can someone help me to prove this inequality : $\left| \sum _{ k=0 }^{ 2n }{ \frac { k }{ k+{ n }^{ 2 } } } -\sum _{ k=0 }^{ 2n }{ \frac { k }{ { n }^{ 2 } } } \right| \le \frac { 4 }{ { n }^{ 2 } } (2n+1)$

Answer 1

First note that $1^2 + 2^2 + 3^2 + ... + n^2 =\frac{n(n + 1)(2n + 1)}{6}$

$\left| \sum _{ k=0 }^{ 2n }{ \frac { k }{ k+{ n }^{ 2 } } } -\sum _{ k=0 }^{ 2n }{ \frac { k }{ { n }^{ 2 } } } \right| \leq \sum _{ k=0 }^{ 2n }\left|\frac { k }{ k+{ n }^{ 2 }} -{ \frac { k }{ { n }^{ 2 } }}\right| = \sum _{ k=0 }^{ 2n }\left| \frac{k^2}{(k+n^2)n^2} \right| \leq \sum _{ k=0 }^{ 2n } \left|\frac{k^2}{n^4}\right| = \frac{1}{n^4}\sum _{ k=0 }^{ 2n } k^2 = \frac{1}{n^4}\frac{2n(2n + 1)(4n + 1)}{6} \leq \frac{1}{n^4}\frac{2n(2n + 1)(12n)}{6} = \frac{4(2n + 1)}{n^2} $

(through this estimation the bound can actually be smaller)