Question Number 34692 by math khazana by abdo last updated on 09/May/18
$${find}\:{lim}_{{n}\rightarrow+\infty} \frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}^{\mathrm{2}} \:{sin}\left(\frac{{k}\pi}{{n}}\right) \\ $$
Commented by math khazana by abdo last updated on 11/May/18
![let put S_n = (1/n^3 ) Σ_(k=1) ^n k^2 sin(((kπ)/n)) S_n = (1/n) Σ_(k=1) ^n ((k/n))^2 sin(((kπ)/n))→_(n→+∞) ∫_0 ^1 x^2 sinx dx let integrate by parts I =∫_0 ^1 x^2 sinxdx =[−x^2 cosx]_0 ^1 +∫_0 ^1 2x cosxdx =−cos(1) +2 { [xsinx]_0 ^1 −∫_0 ^1 sinxdx} = −cos(1) +2{ sin(1) +[cosx]_0 ^1 } =−cos(1) +2{ sin(1) +cos(1)−1} =2sin(1) +cos(1) −2 lim_(n→+∞) S_n = 2sin(1) +cos(1) −2 .](https://www.tinkutara.com/question/Q34828.png)
$${let}\:{put}\:{S}_{{n}} \:=\:\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}^{\mathrm{2}} \:{sin}\left(\frac{{k}\pi}{{n}}\right) \\ $$$${S}_{{n}} =\:\frac{\mathrm{1}}{{n}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\left(\frac{{k}}{{n}}\right)^{\mathrm{2}} \:{sin}\left(\frac{{k}\pi}{{n}}\right)\rightarrow_{{n}\rightarrow+\infty} \:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{2}} \:{sinx}\:{dx} \\ $$$${let}\:{integrate}\:{by}\:{parts} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} {sinxdx}\:=\left[−{x}^{\mathrm{2}} {cosx}\right]_{\mathrm{0}} ^{\mathrm{1}} \:\:\:+\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}{x}\:{cosxdx} \\ $$$$=−{cos}\left(\mathrm{1}\right)\:\:+\mathrm{2}\:\left\{\:\:\left[{xsinx}\right]_{\mathrm{0}} ^{\mathrm{1}} \:\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:{sinxdx}\right\} \\ $$$$=\:−{cos}\left(\mathrm{1}\right)\:\:+\mathrm{2}\left\{\:{sin}\left(\mathrm{1}\right)\:+\left[{cosx}\right]_{\mathrm{0}} ^{\mathrm{1}} \:\right\} \\ $$$$=−{cos}\left(\mathrm{1}\right)\:+\mathrm{2}\left\{\:\:{sin}\left(\mathrm{1}\right)\:+{cos}\left(\mathrm{1}\right)−\mathrm{1}\right\} \\ $$$$=\mathrm{2}{sin}\left(\mathrm{1}\right)\:+{cos}\left(\mathrm{1}\right)\:−\mathrm{2} \\ $$$${lim}_{{n}\rightarrow+\infty} \:\:{S}_{{n}} \:=\:\mathrm{2}{sin}\left(\mathrm{1}\right)\:+{cos}\left(\mathrm{1}\right)\:−\mathrm{2}\:. \\ $$