Question and Answers Forum

All Questions      Topic List

Operation Research Questions

Previous in All Question      Next in All Question      

Previous in Operation Research      Next in Operation Research      

Question Number 66875 by Cmr 237 last updated on 20/Aug/19

Commented by mathmax by abdo last updated on 20/Aug/19

$$convergenceofthisserielet\phi \left(t\right)=\frac{1}{t^3{sin}^{2}t}witht>1$$$$wehave{\phi }^{{}^{\prime }}\left(t\right)=-\frac{3t^2{sin}^{2}t+2t^{3}sintcost}{t^6{sin}^{4}t}\mathrm{\exists }A>0/fort>A$$$${\phi }^{{}^{\prime }}\left(t\right)<0\Rightarrow \phi isdecreazingon]A,+\mathrm{\infty }[so\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^3{sin}^{2}n}$$$$and{\int }_1^{+\mathrm{\infty }}\frac{dt}{t^3{sin}^{2}t}havethesamenature$$$${\int }_1^{+\mathrm{\infty }}\frac{dt}{t^3{sin}^{2}t}thefunctioniscontinueon[1,A[sointegrableon$$$$[1,A[andatV(+\mathrm{\infty }){lim}_{t\to +\mathrm{\infty }}{t}^2\times \frac{1}{t^3{sin}^{2}t}={lim}_{t\to +\mathrm{\infty }}\frac{1}{{tsin}^{2}t}=$$$$\Rightarrow theintegralconverges\Rightarrow thisserieconverges....$$

Commented by Prithwish sen last updated on 20/Aug/19

$$\frac{1}{{\mathrm{t}}^3{\lim }_{\mathrm{t}\to 0}{\left(\frac{\mathrm{sint}}{\mathrm{t}}\right)}^2}=\frac{1}{0.1}\to \mathrm{\infty }$$

Commented by mathmax by abdo last updated on 20/Aug/19

$$nosiriwanttosaythat{lim}_{t\to +\mathrm{\infty }}\frac{1}{t{sin}^{2}t}=0$$

Commented by Prithwish sen last updated on 21/Aug/19

$$\mathrm{oh}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com