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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

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

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