Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

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)

$$\:{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\leqslant{sin}\left({x}\right)\leqslant{x}......{E} \\ $$$$\underset{{k}={n}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\pi}{{k}}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\pi}{{n}+{k}}=\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\pi}{\mathrm{1}+\frac{{k}}{{n}}}={S}_{{n}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{S}_{{n}} =\pi\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{1}+{x}}=\pi{ln}\left(\mathrm{2}\right) \\ $$$$\underset{{k}={n}} {\overset{\mathrm{2}{n}} {\sum}}\left(_{} \frac{\pi}{{n}+{k}}\right)^{\mathrm{3}} ={T}_{{n}} \leqslant\left({n}+\mathrm{1}\right)\left(\frac{\pi}{\mathrm{2}{n}}\right)^{\mathrm{3}} =\frac{\pi^{\mathrm{3}} }{\mathrm{8}}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\right) \\ $$$${E}\Rightarrow \\ $$$${S}_{{n}} −\frac{{T}_{{n}} }{\mathrm{6}}\leqslant\underset{{k}={n}} {\overset{\mathrm{2}{n}} {\sum}}{sin}\left(\frac{\pi}{{k}}\right)\leqslant{S}_{{n}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{S}_{{n}} −\frac{{T}_{{n}} }{\mathrm{6}}\leqslant\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{n}} {\overset{\mathrm{2}{n}} {\sum}}{sin}\left(\frac{\pi}{{k}}\right)\leqslant\underset{{n}\rightarrow\infty} {\mathrm{lim}}{S}_{{n}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}={n}} {\overset{\mathrm{2}{n}} {\sum}}{sin}\left(\frac{\pi}{{k}}\right)=\pi{ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com