Question

5. Use the Limit Comparison Test to determine if the series converges or diverges. n-2 Σ3 -η + 3 n=1

Answer 1

Given Σ3 -η2 + 3 η – 2 n=1

Apply the Sum Rule : an+bna -Σο απ + Σό

$=\sum _{n=1}^{\infty \:}\frac{n}{n^3-n^2+3}-\sum _{n=1}^{\infty \:}\frac{2}{n^3-n^2+3}$

Let us check convergence of each series

$\sum _{n=1}^{\infty \:}\frac{n}{n^3-n^2+3}$

$\mathrm{Apply\:Series\:Comparison\:Test}$

$\mathrm{Let}\:\sum \:a_n,\:\sum \:b_n\:\mathrm{be\:two\:positive\:sequences\:such\:that\:for\:all}\:n,\:\quad \:a_n\le \:b_n$

$\mathrm{If}\:\sum \:b_n\:\mathrm{converges,\:so\:does}\:\sum \:a_n$

If an diverges, so does br.

$\sum _{n=1}^{\infty \:}\frac{n}{n^3-n^2+3}\le \sum _{n=1}^{\infty \:}\frac{n}{n^3-n^2}$

$\mathrm{Check\:convergence\:of\:}\sum _{n=2}^{\infty \:}\frac{n}{n^3-n^2}$

$\mathrm{For\:a\:finite\:upper\:boundary},\:\sum _{n=k}^N\left(a_{n+1}-a_n\right)=a_{N+1}-a_k$

$\mathrm{For\:an\:infinite\:upper\:boundary,\:if}\:a_n\to 0,\:\mathrm{then}\:\sum _{n=k}^{\infty }\left(a_{n+1}-a_n\right)=-a_k$

$\frac{n}{n^3-n^2}=-\left(\frac{1}{n}\right)-\left(-\frac{1}{n-1}\right)$

$a_{n+1}=-\left(\frac{1}{n}\right)\quad \:a_n=-\frac{1}{n-1}\quad \:a_k=a_2=-\frac{1}{2-1}$

$\lim _{n\to \infty \:}\left(-\frac{1}{n-1}\right)=0$

$\mathrm{By\:the\:telescoping\:series\:test:}\quad \sum _{n=2}^{\infty \:}\frac{n}{n^3-n^2}=-a_k$

$=-\left(-\frac{1}{2-1}\right)$

$=1$

$\mathrm{By\:the\:comparison\:test}$

$=\mathrm{converges}$

$\mathrm{Now \:let \:us \:check \:for \:}\sum _{n=1}^{\infty \:}\frac{2}{n^3-n^2+3}$

$=2\cdot \sum _{n=1}^{\infty \:}\frac{1}{n^3-n^2+3}$

$\sum _{n=1}^{\infty \:}\frac{1}{n^3-n^2+3}\le \sum _{n=1}^{\infty \:}\frac{1}{n^3-n^2}$

$\mathrm{Check\:convergence\:of\:}\sum _{n=2}^{\infty \:}\frac{1}{n^3-n^2}$

$\mathrm{Apply\:Series\:Integral\:Test}$

$\mathrm{If\:there\:exists\:an}\:N\ge k\:\mathrm{so\:that\:for\:all}\:n\ge N,\:f\left(n\right)=a_n\:\mathrm{is\:positive,\:continuous\:and\:decreasing}$

$\mathrm{Then}\:\sum _{n=k}^{\infty }a_n\:\mathrm{and}\:\int _k^{\infty }f\left(x\right)dx\:\mathrm{either\:both\:converge\:or\:diverge}$

$\mathrm{Check\:if}\:f\left(n\right)\:\mathrm{is\:positive,\:continuous\:and\:decreasing}$

$\frac{1}{n^3-n^2}\mathrm{\:is\:positive,\:continuous\:and\:decreasing\:from\:}n=2$

$\int _2^{\infty \:}\frac{1}{n^3-n^2}dn=-\left(\frac{1}{2}-\ln \left(2\right)\right)$

$\mathrm{By\:the\:integral\:test\:criteria}$

$=\mathrm{converges}$

$\mathrm{By\:the\:comparison\:test}$

$=\mathrm{converges}$

$\mathrm{Hence\:}\Rightarrow \mathrm{converges}-\mathrm{converges}$

$=\boxed{\mathrm{converges}}$