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Question Number 157883 by HongKing last updated on 29/Oct/21

lim_(n→∞) ((1/(n^2 +1)) + (2/(n^2 +1)) + ... + ((n-1)/(n^2 +1))) = ?

$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}\:+\:\frac{\mathrm{2}}{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}\:+\:...\:+\:\frac{\mathrm{n}-\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}\right)\:=\:? \\ $$

Answered by puissant last updated on 29/Oct/21

S=lim_(n→∞) ((1/(n^2 +1))+(2/(n^2 +1))+...+((n−1)/(n^2 +1))) = lim_(n→∞) (1/(n^2 +1)) Σ_(k=1) ^(n−1) k =lim_(n→∞) (((n−1)n)/(2(n^2 +1))) = (1/2)lim_(n→∞) ((n^2 (1−(1/n)))/(n^2 (1+(1/n^2 )))) =(1/2)lim_(n→∞) (({1−(1/n)})/({1+(1/n^2 )}))= (1/2).. ............Le puissant...........

$${S}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{2}}{{n}^{\mathrm{2}} +\mathrm{1}}+...+\frac{{n}−\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}\:\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{k}\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}\left({n}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)}{{n}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)}\:=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\left\{\mathrm{1}−\frac{\mathrm{1}}{{n}}\right\}}{\left\{\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right\}}=\:\frac{\mathrm{1}}{\mathrm{2}}.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:............\mathscr{L}{e}\:{puissant}........... \\ $$

Commented by Tawa11 last updated on 29/Oct/21

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by HongKing last updated on 29/Oct/21

alot thanks sir

$$\mathrm{alot}\:\mathrm{thanks}\:\mathrm{sir} \\ $$

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