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Question Number 160706 by HongKing last updated on 05/Dec/21

Answered by mr W last updated on 05/Dec/21

$$S_n=\underset{k=1}{\sum ^n}\sin \left(\frac{k}{n^3}\right)$$$$T_n=\underset{k=1}{\sum ^n}\cos \left(\frac{k}{n^3}\right)$$$$T_n+{iS}_n=\underset{k=1}{\sum ^n}\{\cos \left(\frac{k}{n^3}\right)+i\sin \left(\frac{k}{n^3}\right)\}$$$$T_n+{iS}_n=\underset{k=1}{\sum ^n}{e}^{\frac{ik}{n^3}}=\frac{e^{\frac{i}{n^3}}(e^{\frac{in}{n^3}}-1)}{e^{\frac{i}{n^3}}-1}$$$$T_n+{iS}_n=\frac{(\cos \frac{1}{n^3}+i\sin \frac{1}{n^3})[(\cos \frac{1}{n^2}-1)+i\sin \frac{1}{n^2}]}{(\cos \frac{1}{n^3}-1)+i\sin \frac{1}{n^3}}$$$$T_n+{iS}_n=\frac{[\cos \frac{1}{n^3}(\cos \frac{1}{n^2}-1)-\sin \frac{1}{n^3}\sin \frac{1}{n^2}]+i[\cos \frac{1}{n^3}\sin \frac{1}{n^2}+\sin \frac{1}{n^3}(\cos \frac{1}{n^2}-1)]}{(\cos \frac{1}{n^3}-1)+i\sin \frac{1}{n^3}}$$$$T_n+{iS}_n=\frac{[\cos (\frac{1}{n^3}+\frac{1}{n^2})-\cos \frac{1}{n^3}]+i[\sin (\frac{1}{n^3}+\frac{1}{n^2})-\sin \frac{1}{n^3}]}{(\cos \frac{1}{n^3}-1)+i\sin \frac{1}{n^3}}$$$$T_n+{iS}_n=\frac{\{[\cos (\frac{1}{n^3}+\frac{1}{n^2})-\cos \frac{1}{n^3}]+i[\sin (\frac{1}{n^3}+\frac{1}{n^2})-\sin \frac{1}{n^3}]\}[(\cos \frac{1}{n^3}-1)-i\sin \frac{1}{n^3}]}{{(\cos \frac{1}{n^3}-1)}^2+{\left(\sin \frac{1}{n^3}\right)}^2}$$$$T_n+{iS}_n=\frac{\{(\cos \frac{n+1}{n^3}-\cos \frac{1}{n^3})(\cos \frac{1}{n^3}-1)+\sin \frac{1}{n^3}(\sin \frac{n+1}{n^3}-\sin \frac{1}{n^3})\}+i\{(\cos \frac{1}{n^3}-1)(\sin \frac{n+1}{n^3}-\sin \frac{1}{n^3})-\sin \frac{1}{n^3}(\cos \frac{n+1}{n^3}-\cos \frac{1}{n^3})\}}{{(\cos \frac{1}{n^3}-1)}^2+{\left(\sin \frac{1}{n^3}\right)}^2}$$$$\Rightarrow T_n=\frac{(\cos \frac{n+1}{n^3}-\cos \frac{1}{n^3})(\cos \frac{1}{n^3}-1)+\sin \frac{1}{n^3}(\sin \frac{n+1}{n^3}-\sin \frac{1}{n^3})}{{(\cos \frac{1}{n^3}-1)}^2+{\left(\sin \frac{1}{n^3}\right)}^2}$$$$\Rightarrow T_n=\frac{1}{2}\{\frac{\cos \frac{1}{n^2}-\cos (\frac{1}{n^3}+\frac{1}{n^2})\cos \frac{1}{n^3}}{1-\cos \frac{1}{n^3}}-1\}$$$$\Rightarrow T_n=\frac{1}{2}\{\frac{\sin \frac{1}{n^3}\sin (\frac{1}{n^3}+\frac{1}{n^2})}{1-\cos \frac{1}{n^3}}-1\}$$$$\Rightarrow S_n=\frac{(\cos \frac{1}{n^3}-1)(\sin \frac{n+1}{n^3}-\sin \frac{1}{n^3})-\sin \frac{1}{n^3}(\cos \frac{n+1}{n^3}-\cos \frac{1}{n^3})}{{(\cos \frac{1}{n^3}-1)}^2+{\left(\sin \frac{1}{n^3}\right)}^2}$$$$\Rightarrow S_n=\frac{\sin \frac{1}{n^3}+\sin \frac{1}{n^2}-\sin (\frac{1}{n^3}+\frac{1}{n^2})}{2(1-\cos \frac{1}{n^3})}$$$$\Rightarrow S_n=\frac{1}{2}\{\frac{1-\cos \frac{1}{n^2}}{1-\cos \frac{1}{n^3}}\times \sin \frac{1}{n^3}+\sin \frac{1}{n^2}\}$$$$\underset{n\to \mathrm{\infty }}{\lim }{S}_n=\frac{1}{2}\underset{n\to \mathrm{\infty }}{\lim }\{\frac{1-\cos \frac{1}{n^2}}{1-\cos \frac{1}{n^3}}\times \sin \frac{1}{n^3}+\sin \frac{1}{n^2}\}$$$$=\frac{1}{2}\underset{x\to 0}{\lim }\{\frac{1-\cos x^2}{1-\cos x^3}\times \sin x^3+\sin x^2\}$$$$=\frac{1}{2}\underset{x\to 0}{\lim }\{0+0\}$$$$=0$$$$$$$$\underset{x\to 0}{\lim }\left\{\frac{1-\cos x^2}{1-\cos x^3}\times \sin x^3\right\}$$$$=\underset{x\to 0}{\lim }\left\{\frac{\frac{x^4}{2}-\frac{x^8}{24}+...}{\frac{x^6}{2}-\frac{x^{12}}{24}+...}\times (x^3-\frac{x^6}{6}+...)\right\}$$$$=\underset{x\to 0}{\lim }\left\{\frac{x(\frac{1}{2}-\frac{x^4}{24}+...)}{\frac{1}{2}-\frac{x^6}{24}+...}\times (1-\frac{x^3}{6}+...)\right\}$$$$=0$$

Answered by mindispower last updated on 05/Dec/21

$$0⩽sin\left(x\right)⩽x,\mathrm{\forall }x\in [0,\frac{\pi }{2}]$$$$proof,cos\left(x\right)⩽1\Rightarrow {\int }_0^{t}cos\left(x\right)dx⩽{\int }_0^{t}dx$$$$\iff sin\left(t\right)⩽t$$$$\mathrm{\forall }k\in [1,n],\frac{k}{n^3}\in [0,1[\subset [0,\frac{\pi }{2}]$$$$0⩽S_n⩽\underset{k=1}{\sum ^n}\frac{k}{n^3}=\frac{1}{n^3}.\frac{n(1+n)}{2}$$$$\Rightarrow {LimS}_n=0$$$$$$

Commented by mr W last updated on 05/Dec/21

$$nicemethodforthelimit!$$$$canyoupleasecomfirmifmy$$$$formulasfor{S}_{n}and{T}_{n}arecorrect?$$

Commented by mindispower last updated on 05/Dec/21

$$thankYou$$$$nicewayyougothrotoo$$$$correctwaysirniceday$$

Commented by mr W last updated on 06/Dec/21

$$thankssir!$$

Answered by qaz last updated on 05/Dec/21

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\sin \left(\frac{\mathrm{k}}{{\mathrm{n}}^3}\right)=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}(\frac{\mathrm{k}}{{\mathrm{n}}^3}+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^3}\right))=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }(\frac{\mathrm{n}+1}{{2\mathrm{n}}^2}+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^2}\right))=0$$

Answered by mathmax by abdo last updated on 06/Dec/21

$${\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\sin \left(\frac{\mathrm{k}}{{\mathrm{n}}^3}\right)\mathrm{we}\mathrm{know}\mathrm{x}-\frac{{\mathrm{x}}^3}{6}⩽\mathrm{sinx}⩽\mathrm{x}\Rightarrow$$$$\frac{\mathrm{k}}{{\mathrm{n}}^3}-\frac{1}{6}\frac{{\mathrm{k}}^3}{{\mathrm{n}}^9}⩽\sin \left(\frac{\mathrm{k}}{{\mathrm{n}}^3}\right)⩽\frac{\mathrm{k}}{{\mathrm{n}}^3}\Rightarrow$$$$\frac{1}{{\mathrm{n}}^3}\sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{k}-\frac{1}{{6\mathrm{n}}^9}\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{k}}^3⩽\sum _{\mathrm{k}=1}^{\mathrm{n}}\sin \left(\frac{\mathrm{k}}{{\mathrm{n}}^3}\right)⩽\frac{1}{{\mathrm{n}}^3}\sum _{\mathrm{i}=1}^{\mathrm{n}}\mathrm{k}\Rightarrow$$$$\frac{\mathrm{n}(\mathrm{n}+1)}{{2\mathrm{n}}^3}-\frac{1}{{6\mathrm{n}}^9}\frac{{\mathrm{n}}^2{(\mathrm{n}+1)}^2}{4}⩽{\mathrm{S}}_{\mathrm{n}}⩽\frac{\mathrm{n}(\mathrm{n}+1)}{{2\mathrm{n}}^3}\Rightarrow$$$$\frac{\mathrm{n}+1}{{2\mathrm{n}}^2}-\frac{{(\mathrm{n}+1)}^2}{{24\mathrm{n}}^7}(\to 0)⩽{\mathrm{S}}_{\mathrm{n}}⩽\frac{\mathrm{n}+1}{{2\mathrm{n}}^2}(\to 0)\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{S}}_{\mathrm{n}}=0$$$$$$

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