Question Number 144396 by SOMEDAVONG last updated on 25/Jun/21
$$\mathrm{L}=\underset{\mathrm{n}\rightarrow+\propto} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }\:+\:\frac{\mathrm{2}}{\mathrm{n}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }\:+..+\:\frac{\mathrm{n}}{\mathrm{n}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} }\right)=? \\ $$
Commented by Canebulok last updated on 25/Jun/21
![Solution: ⇒ lim_(n→+∞) (Σ_(k=1) ^n (k/(n^2 +k^2 )) ) = L ⇒ lim_(n→+∞) ((1/n)) (Σ_(k=1) ^n ((nk)/(n^2 +k^2 )) ) = L ⇒ lim_(n→+∞) ((1/n)) (Σ_(k=1) ^n (1/((n/k)+(k/n))) ) = L ∴ ⇒ ∫_0 ^( 1) (1/(((1/x) + x))) dx = L ⇒ ∫_0 ^( 1) (x/(x^2 +1)) dx = [((Ln(∣x^2 +1∣))/2) ∣_0 ^1 ] ⇒ L = ((Ln(∣2∣))/2)](https://www.tinkutara.com/question/Q144398.png)
$$\: \\ $$$$\boldsymbol{{Solution}}: \\ $$$$\Rightarrow\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{{k}}{{n}^{\mathrm{2}} +{k}^{\mathrm{2}} }\:\right)\:=\:{L} \\ $$$$\Rightarrow\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{{n}}\right)\:\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{{nk}}{{n}^{\mathrm{2}} +{k}^{\mathrm{2}} }\:\right)\:=\:{L} \\ $$$$\Rightarrow\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{{n}}\right)\:\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{\frac{{n}}{{k}}+\frac{{k}}{{n}}}\:\right)\:=\:{L} \\ $$$$\therefore \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{{x}}\:+\:{x}\right)}\:{dx}\:=\:{L} \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}\:=\:\left[\frac{{Ln}\left(\mid{x}^{\mathrm{2}} +\mathrm{1}\mid\right)}{\mathrm{2}}\:\mid_{\mathrm{0}} ^{\mathrm{1}} \:\right] \\ $$$$\Rightarrow\:{L}\:=\:\frac{{Ln}\left(\mid\mathrm{2}\mid\right)}{\mathrm{2}} \\ $$
Answered by mathmax by abdo last updated on 25/Jun/21
![L=lim_(n→+∞) Σ_(k=1) ^n (k/(n^2 +k^2 )) =lim_(n→+∞) ((k/n)/(n+(k^2 /n))) =lim_(n→+∞) (1/n)×((k/n)/(1+((k/n))^2 ))=∫_0 ^1 (x/(1+x^2 ))dx =(1/2)[log(1+x^2 )]_0 ^1 =(1/2)log2](https://www.tinkutara.com/question/Q144448.png)
$$\mathrm{L}=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{k}}{\mathrm{n}^{\mathrm{2}} \:+\mathrm{k}^{\mathrm{2}} }\:=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\:\:\:\frac{\frac{\mathrm{k}}{\mathrm{n}}}{\mathrm{n}+\frac{\mathrm{k}^{\mathrm{2}} }{\mathrm{n}}} \\ $$$$=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \frac{\mathrm{1}}{\mathrm{n}}×\frac{\frac{\mathrm{k}}{\mathrm{n}}}{\mathrm{1}+\left(\frac{\mathrm{k}}{\mathrm{n}}\right)^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{log}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log2} \\ $$