Question and Answers Forum

All Questions      Topic List

Number Theory Questions

Previous in All Question      Next in All Question      

Previous in Number Theory      Next in Number Theory      

Question Number 140833 by liberty last updated on 13/May/21

$$\mathrm{Determine}\mathrm{if}\mathrm{the}\mathrm{series}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{\mathrm{a}}_{\mathrm{n}}$$$$\mathrm{defined}\mathrm{by}\mathrm{the}\mathrm{formula}\mathrm{converges}\mathrm{or}$$$$\mathrm{diverges}.{\mathrm{a}}_1=4,{\mathrm{a}}_{\mathrm{n}+1}=\frac{10+\sin \mathrm{n}}{\mathrm{n}}.{\mathrm{a}}_{\mathrm{n}}$$

Commented by liberty last updated on 13/May/21

can anyone help me

Answered by TheSupreme last updated on 13/May/21

$$\frac{a_{n+1}}{a_n}=\frac{10+sin\left(n\right)}{n}$$$$lim\frac{a_{n+1}}{a_n}=0\to seriesconverges$$

Commented by liberty last updated on 13/May/21

$$\mathrm{why}\lim \frac{{\mathrm{a}}_{\mathrm{n}+1}}{{\mathrm{a}}_{\mathrm{n}}}=0?\mathrm{as}\mathrm{n}\to ?$$

Answered by mathmax by abdo last updated on 13/May/21

$${\mathrm{a}}_{\mathrm{n}+1}=\frac{10+\sin \left(\mathrm{n}\right)}{\mathrm{n}}{\mathrm{a}}_{\mathrm{n}}\Rightarrow \frac{{\mathrm{a}}_{\mathrm{n}+1}}{{\mathrm{a}}_{\mathrm{n}}}=\frac{10+\sin \left(\mathrm{n}\right)}{\mathrm{n}}\Rightarrow$$$$\prod _{\mathrm{k}=1}^{\mathrm{n}-1}\frac{{\mathrm{a}}_{\mathrm{k}+1}}{{\mathrm{a}}_{\mathrm{k}}}=\prod _{\mathrm{k}=1}^{\mathrm{n}-1}\left(\frac{10+\sin \left(\mathrm{k}\right)}{\mathrm{k}}\right)\Rightarrow \frac{{\mathrm{a}}_2}{{\mathrm{a}}_1}.\frac{{\mathrm{a}}_3}{{\mathrm{a}}_2}.....\frac{{\mathrm{a}}_{\mathrm{n}}}{{\mathrm{a}}_{\mathrm{n}-1}}=\frac{1}{(\mathrm{n}-1)!}\prod _{\mathrm{k}=1}^{\mathrm{n}-1}(10+\sin \left(\mathrm{k}\right))\Rightarrow$$$${\mathrm{a}}_{\mathrm{n}}=\frac{4}{(\mathrm{n}-1)!}\prod _{\mathrm{k}=1}^{\mathrm{n}-1}(10+\sin \left(\mathrm{k}\right))\Rightarrow \mid {\mathrm{a}}_{\mathrm{n}}\mid ⩽\frac{4.{11}^{\mathrm{n}}}{(\mathrm{n}-1)!}={\mathrm{v}}_{\mathrm{n}}$$$$\frac{{\mathrm{v}}_{\mathrm{n}+1}}{{\mathrm{v}}_{\mathrm{n}}}=\frac{4.{11}^{\mathrm{n}+1}}{\mathrm{n}!}\times \frac{(\mathrm{n}-1)!}{4.{11}^{\mathrm{n}}}=\frac{11}{\mathrm{n}}\to 0(\mathrm{n}\to \mathrm{\infty })\Rightarrow \mathrm{\Sigma }{\mathrm{v}}_{\mathrm{n}}\mathrm{converges}\Rightarrow$$$$\mathrm{\Sigma }{\mathrm{a}}_{\mathrm{n}}\mathrm{cv}.$$

Answered by physicstutes last updated on 13/May/21

$$a_{n+1}=\frac{10+\sin n}{n}{a}_n$$$$\Rightarrow \frac{a_{n+1}}{a_n}=\frac{10+\sin n}{n}=\frac{10}{n}+\frac{\sin n}{n}$$$$\underset{n\to \mathrm{\infty }}{\lim }\frac{a_{n+1}}{a_n}=\underset{n\to \mathrm{\infty }}{\lim }(\frac{10}{n}+\frac{\sin n}{n})=0+0=0<1$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}{a}_n\mathrm{converges}\mathrm{by}\mathrm{the}\mathrm{ratio}\mathrm{test}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com