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Question Number 124355 by Eric002 last updated on 02/Dec/20

find the limit lim_(n→∞) (sin((1/n^2 ))+sin((2/n^2 ))+......+sin((n/n^2 )))

$${find}\:{the}\:{limit}\: \\ $$$$\underset{{n}\rightarrow\infty} {{lim}}\left({sin}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)+{sin}\left(\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\right)+......+{sin}\left(\frac{{n}}{{n}^{\mathrm{2}} }\right)\right) \\ $$

Answered by Dwaipayan Shikari last updated on 02/Dec/20

lim_(n→∞) sin((1/n^2 ))+sin((2/n^2 ))+... =((1+2+3+4+5+6+..+n)/n^2 )=((n^2 +n)/(2n^2 ))=(1/2) (As sin((1/n^2 ))→((1/n^2 )))

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{sin}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)+{sin}\left(\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\right)+... \\ $$$$=\frac{\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+..+{n}}{{n}^{\mathrm{2}} }=\frac{{n}^{\mathrm{2}} +{n}}{\mathrm{2}{n}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\left({As}\:{sin}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)\rightarrow\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)\right) \\ $$

Answered by mathmax by abdo last updated on 02/Dec/20

we have x−(x^3 /6)≤sinx≤x ⇒Σ_(k=1) ^n (k/n^2 )−(1/6)Σ_(k=1) ^n (k^3 /n^6 )≤Σ_(k=1) ^n sin((k/n^2 ))≤Σ_(k=1) ^n (k/n^2 ) ⇒ ((n(n+1))/(2n^2 ))−(1/(6n^3 ))(((n(n+1))/2))^2 ≤ S_n ≤((n(n+1))/(2n^2 )) we passe to[limit (n→∞) lim_(n→+∞) S_n =(1/2)

$$\mathrm{we}\:\mathrm{have}\:\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\leqslant\mathrm{sinx}\leqslant\mathrm{x}\:\Rightarrow\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{k}}{\mathrm{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{6}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{k}^{\mathrm{3}} }{\mathrm{n}^{\mathrm{6}} }\leqslant\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{sin}\left(\frac{\mathrm{k}}{\mathrm{n}^{\mathrm{2}} }\right)\leqslant\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{k}}{\mathrm{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{6n}^{\mathrm{3}} }\left(\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}\right)^{\mathrm{2}} \:\leqslant\:\mathrm{S}_{\mathrm{n}} \leqslant\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2n}^{\mathrm{2}} }\:\:\:\mathrm{we}\:\mathrm{passe}\:\mathrm{to}\left[\mathrm{limit}\:\left(\mathrm{n}\rightarrow\infty\right)\right. \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{S}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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