Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 120978 by ZiYangLee last updated on 04/Nov/20

lim_(x→∞) ((1/(n^2 +π))+(1/(n^2 +2π))+…+(1/(n^2 +nπ)))=?

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\pi}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{2}\pi}+\ldots+\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}\pi}\right)=? \\ $$

Commented by Dwaipayan Shikari last updated on 04/Nov/20

If the question becomes  lim_(n→∞) ((1/(n+π))+(1/(n+2π))+....+(1/(n+nπ)))  lim_(n→∞) (1/n)Σ_(k=1) ^∞ (1/(1+((kπ)/n)))  =∫_0 ^1 (1/(1+πx))dx=(1/π)∫_0 ^π (du/(1+u))=(1/π)[log(1+u)]_0 ^π =(1/π)log(1+π)

$${If}\:{the}\:{question}\:{becomes} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{n}+\pi}+\frac{\mathrm{1}}{{n}+\mathrm{2}\pi}+....+\frac{\mathrm{1}}{{n}+{n}\pi}\right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{1}+\frac{{k}\pi}{{n}}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+\pi{x}}{dx}=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} \frac{{du}}{\mathrm{1}+{u}}=\frac{\mathrm{1}}{\pi}\left[{log}\left(\mathrm{1}+{u}\right)\right]_{\mathrm{0}} ^{\pi} =\frac{\mathrm{1}}{\pi}{log}\left(\mathrm{1}+\pi\right) \\ $$

Answered by Olaf last updated on 04/Nov/20

n^2 +kπ ≥ n^2 , k∈N^∗   (1/(n^2 +kπ)) ≤ (1/n^2 )  0 ≤ Σ_(k=1) ^n (1/(n^2 +kπ)) ≤ Σ_(k=1) ^n (1/n^2 ) = n×(1/n^2 ) = (1/n)  lim_(n→∞) (1/n) = 0  ⇒ lim_(n→∞) Σ_(k=1) ^n (1/(n^2 +kπ)) = 0

$${n}^{\mathrm{2}} +{k}\pi\:\geqslant\:{n}^{\mathrm{2}} ,\:{k}\in\mathbb{N}^{\ast} \\ $$$$\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{k}\pi}\:\leqslant\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\mathrm{0}\:\leqslant\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{k}\pi}\:\leqslant\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\:{n}×\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{{n}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{k}\pi}\:=\:\mathrm{0} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com