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lim-n-k-n-2n-sin-pi-k-




Question Number 166699 by qaz last updated on 25/Feb/22
lim_(n→∞) Σ_(k=n) ^(2n) sin (π/k)=?
$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\underset{\mathrm{k}=\mathrm{n}} {\overset{\mathrm{2n}} {\sum}}\mathrm{sin}\:\frac{\pi}{\mathrm{k}}=? \\ $$
Answered by mathsmine last updated on 25/Feb/22
 x−(x^3 /6)≤sin(x)≤x......E  Σ_(k=n) ^(2n) (π/k)=Σ_(k=0) ^n (π/(n+k))=(1/n)Σ_(k=0) ^n (π/(1+(k/n)))=S_n   lim_(n→∞) S_n =π∫_0 ^1 (dx/(1+x))=πln(2)  Σ_(k=n) ^(2n) (_ (π/(n+k)))^3 =T_n ≤(n+1)((π/(2n)))^3 =(π^3 /8)((1/n^2 )+(1/n^3 ))  E⇒  S_n −(T_n /6)≤Σ_(k=n) ^(2n) sin((π/k))≤S_n   lim_(n→∞) S_n −(T_n /6)≤lim_(n→∞) Σ_n ^(2n) sin((π/k))≤lim_(n→∞) S_n   lim_(n→∞) Σ_(k=n) ^(2n) sin((π/k))=πln(2)
$$\:{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\leqslant{sin}\left({x}\right)\leqslant{x}……{E} \\ $$$$\underset{{k}={n}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\pi}{{k}}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\pi}{{n}+{k}}=\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\pi}{\mathrm{1}+\frac{{k}}{{n}}}={S}_{{n}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{S}_{{n}} =\pi\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{1}+{x}}=\pi{ln}\left(\mathrm{2}\right) \\ $$$$\underset{{k}={n}} {\overset{\mathrm{2}{n}} {\sum}}\left(_{} \frac{\pi}{{n}+{k}}\right)^{\mathrm{3}} ={T}_{{n}} \leqslant\left({n}+\mathrm{1}\right)\left(\frac{\pi}{\mathrm{2}{n}}\right)^{\mathrm{3}} =\frac{\pi^{\mathrm{3}} }{\mathrm{8}}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\right) \\ $$$${E}\Rightarrow \\ $$$${S}_{{n}} −\frac{{T}_{{n}} }{\mathrm{6}}\leqslant\underset{{k}={n}} {\overset{\mathrm{2}{n}} {\sum}}{sin}\left(\frac{\pi}{{k}}\right)\leqslant{S}_{{n}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{S}_{{n}} −\frac{{T}_{{n}} }{\mathrm{6}}\leqslant\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{n}} {\overset{\mathrm{2}{n}} {\sum}}{sin}\left(\frac{\pi}{{k}}\right)\leqslant\underset{{n}\rightarrow\infty} {\mathrm{lim}}{S}_{{n}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}={n}} {\overset{\mathrm{2}{n}} {\sum}}{sin}\left(\frac{\pi}{{k}}\right)=\pi{ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$

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