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Question Number 212595 by MrGaster last updated on 18/Oct/24

$$\underset{n\to \mathrm{\infty }}{\lim }(\frac{1^2}{n^2+1}+\frac{2^2}{n^2+2}+\frac{3^2}{n^2+3}+\dots +\frac{n^2}{n^2+n}-\frac{n}{3})=?$$

Answered by Berbere last updated on 18/Oct/24

$$=\mathrm{\Sigma }\frac{k^2}{n^2+n}⩽\mathrm{\Sigma }\frac{k^2}{n^2+k}-\frac{n}{3}⩽\underset{k=1}{\sum ^n}\frac{k^2}{n^2}-\frac{n}{3}$$$$U_n=\frac{n(2n+1)(n+1)}{6(n^2+n)}-\frac{n}{3}⩽S_n⩽\frac{1}{n^2}(2n+1)n(n+1)-\frac{n}{3}=V_n$$$$U_n\to 0;v_n\to 0\Rightarrow S_n\to 0$$$$$$

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