# If-n-1-a-n-is-convergent-and-n-1-b-n-is-convergent-What-are-the-sufficient-condition-so-that-n-1-a-n-n-b-n-1-n-converges-a-n-b-n-gt-0-

Question Number 3460 by prakash jain last updated on 14/Dec/15
$$\mathrm{If}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} \:\mathrm{is}\:\mathrm{convergent}\:\mathrm{and}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{b}_{{n}} \:\mathrm{is}\:\mathrm{convergent}. \\$$$$\mathrm{What}\:\mathrm{are}\:\mathrm{the}\:\mathrm{sufficient}\:\mathrm{condition}\:\mathrm{so}\:\mathrm{that} \\$$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} \:\mathrm{converges}. \\$$$${a}_{{n}} ,\:{b}_{{n}} >\mathrm{0} \\$$
Commented by Yozzii last updated on 14/Dec/15
$${Let}\:{s}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} \\$$$$\Rightarrow{e}^{{s}} ={e}^{\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} } \\$$$${What}\:{is}\:{b}_{\mathrm{0}} ^{\mathrm{1}/\mathrm{0}} \:? \\$$
Commented by prakash jain last updated on 13/Dec/15
$${a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} =\left({a}_{{n}} \right)^{{n}} ×\left({b}_{{n}} \right)^{\mathrm{1}/{n}} ,\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\mathrm{0}/\mathrm{0}? \\$$$$\\$$
Commented by prakash jain last updated on 14/Dec/15
$$\mathrm{Question}\:\mathrm{correctd}\:{n}\:\mathrm{should}\:\mathrm{start}\:\mathrm{from}\:\mathrm{1}. \\$$
Commented by Filup last updated on 13/Dec/15
$$\mathrm{if}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{b}_{{n}} =\mathrm{0} \\$$$$\mathrm{Doesn}'\mathrm{t}\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{a}_{{n}} }{{b}_{{n}} }\:=\:\frac{\mathrm{0}}{\mathrm{0}}? \\$$$$\\$$$$\therefore{must}\:{be}\:{differentiable}? \\$$
Commented by Filup last updated on 14/Dec/15
$${sorry},\:{i}\:{missread}\:{b}_{{n}} ^{\mathrm{1}/{n}} \:{as}\:{b}_{{n}} ^{−\mathrm{1}} \\$$$$\\$$$${and}\:{i}\:{was}\:{reffering}\:{to}\:{how}\:{if} \\$$$${x}_{{n}} \:{converges},\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{x}_{{n}} =\mathrm{0}.\:\mathrm{if}\:\mathrm{i}\:\mathrm{am}\:\mathrm{not}\:\mathrm{mistaken} \\$$
Commented by 123456 last updated on 14/Dec/15
$$\mathrm{if}\:\mathrm{im}\:\mathrm{not}\:\mathrm{wrong}. \\$$$$\mathrm{we}\:\mathrm{asking}\:\mathrm{for}\:\mathrm{sulficient}\:\mathrm{conditions}, \\$$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} =\mathrm{0}\:\mathrm{is}\:\mathrm{only}\:\mathrm{a}\:\mathrm{necessary},\:\mathrm{but}\:\mathrm{not} \\$$$$\mathrm{sulficient}\:\left({a}_{{n}} =\mathrm{1}/{n}\:\mathrm{is}\:\mathrm{any}\:\mathrm{example}\right. \\$$$$\left.{a}_{{n}} \rightarrow\mathrm{0}\:\mathrm{as}\:{n}\rightarrow\infty,\:\mathrm{but}\:\Sigma{a}_{{n}} \:\mathrm{diverged}\right) \\$$
Commented by Yozzii last updated on 14/Dec/15
$${s}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} \\$$$${e}^{{s}} ={e}^{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} } \\$$$${e}^{{s}} ={e}^{{a}_{\mathrm{1}} {b}_{\mathrm{1}} } {e}^{{a}_{\mathrm{2}} ^{\mathrm{2}} {b}_{\mathrm{2}} ^{\mathrm{1}/\mathrm{2}} } {e}^{{a}_{\mathrm{3}} ^{\mathrm{3}} {b}_{\mathrm{3}} ^{\mathrm{1}/\mathrm{3}} } {e}^{{a}_{\mathrm{4}} ^{\mathrm{4}} {b}_{\mathrm{4}} ^{\mathrm{1}/\mathrm{4}} } …. \\$$$${e}^{{s}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left\{{e}^{{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} } −\mathrm{1}+\mathrm{1}\right\} \\$$$${let}\:{g}_{{n}} ={e}^{{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} } −\mathrm{1} \\$$$$\therefore\:{e}^{{s}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left({g}_{{n}} +\mathrm{1}\right) \\$$$${if}\:{e}^{{s}} \:{is}\:{defined},\:{the}\:{infinite}\:{product} \\$$$${on}\:{the}\:{rhs}\:{converges}.\:{If}\:{this}\:{is}\:{so}, \\$$$${then}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{g}_{{n}} \:{converges}.\: \\$$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({e}^{{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} } −\mathrm{1}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{e}^{{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} } −\infty\: \\$$$${If}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{g}_{{n}} =\mathrm{0} \\$$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{e}^{{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} } =\mathrm{1} \\$$$$\\$$$$\\$$$$\\$$$$\\$$$$\\$$
Commented by prakash jain last updated on 14/Dec/15
$$\mathrm{Yes}.\:\mathrm{The}\:\mathrm{question}\:\mathrm{is}\:\mathrm{asking}\:\mathrm{for}\:\mathrm{sufficient} \\$$$$\mathrm{condition}. \\$$
Commented by Yozzii last updated on 14/Dec/15
$${Yes}… \\$$
Commented by prakash jain last updated on 14/Dec/15
$$\underset{{n}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} } =\mathrm{1}\:{if} \\$$$$\underset{{n}\rightarrow\mathrm{0}} {\mathrm{lim}}{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} =\mathrm{0} \\$$$$\mathrm{Since}\:{c}_{{n}} ={a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} \:\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{necessary}\:\mathrm{condition} \\$$$$\mathrm{that}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{c}_{{n}} =\mathrm{0},\:\mathrm{but}\:\mathrm{this}\:\mathrm{is}\:\mathrm{not}\:\mathrm{sufficient}. \\$$