Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 100539 by Dwaipayan Shikari last updated on 27/Jun/20

$${(-1)}^n\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{3^n}{n}$$

Commented by maths mind last updated on 27/Jun/20

$$\sum _{n⩾1}{(-1)}^n\frac{3^n}{n}thisdiverge$$

Commented by abdomathmax last updated on 27/Jun/20

$$\mathrm{S}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{u}}_{\mathrm{n}}\mathrm{with}{\mathrm{u}}_{\mathrm{n}}=\frac{{(-3)}^{\mathrm{n}}}{\mathrm{n}}\mathrm{we}\mathrm{have}$$$$\mid \frac{{\mathrm{u}}_{\mathrm{n}+1}}{{\mathrm{u}}_{\mathrm{n}}}\mid =\frac{3^{\mathrm{n}+1}}{\mathrm{n}+1}\times \frac{\mathrm{n}}{3^{\mathrm{n}}}=\frac{3\mathrm{n}}{\mathrm{n}+1}\to 3>1\Rightarrow \mathrm{\Sigma }{\mathrm{u}}_{\mathrm{n}}\mathrm{diverges}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com