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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-

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Question Number 3460 by prakash jain last updated on 14/Dec/15
If Σ_(n=1) ^∞ a_n is convergent and Σ_(n=1) ^∞ b_n is convergent. What are the sufficient condition so that Σ_(n=1) ^∞ a_n ^n b_n ^(1/n) converges. a_n , b_n >0
$$\mathrm{If}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} \:\mathrm{is}\:\mathrm{convergent}\:\mathrm{and}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{b}_{{n}} \:\mathrm{is}\:\mathrm{convergent}. \\ $$$$\mathrm{What}\:\mathrm{are}\:\mathrm{the}\:\mathrm{sufficient}\:\mathrm{condition}\:\mathrm{so}\:\mathrm{that} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} \:\mathrm{converges}. \\ $$$${a}_{{n}} ,\:{b}_{{n}} >\mathrm{0} \\ $$
Commented by Yozzii last updated on 14/Dec/15
Let s=Σ_(n=0) ^∞ a_n ^n b_n ^(1/n) ⇒e^s =e^(Σ_(n=0) ^∞ a_n ^n b_n ^(1/n) ) What is b_0 ^(1/0) ?
$${Let}\:{s}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} \\ $$$$\Rightarrow{e}^{{s}} ={e}^{\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} } \\ $$$${What}\:{is}\:{b}_{\mathrm{0}} ^{\mathrm{1}/\mathrm{0}} \:? \\ $$
Commented by prakash jain last updated on 13/Dec/15
a_n ^n b_n ^(1/n) =(a_n )^n ×(b_n )^(1/n) , how did you get 0/0?
$${a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} =\left({a}_{{n}} \right)^{{n}} ×\left({b}_{{n}} \right)^{\mathrm{1}/{n}} ,\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\mathrm{0}/\mathrm{0}? \\ $$$$ \\ $$
Commented by prakash jain last updated on 14/Dec/15
Question correctd n should start from 1.
$$\mathrm{Question}\:\mathrm{correctd}\:{n}\:\mathrm{should}\:\mathrm{start}\:\mathrm{from}\:\mathrm{1}. \\ $$
Commented by Filup last updated on 13/Dec/15
if lim_(n→∞) b_n =0 Doesn′t lim_(n→∞) (a_n /b_n ) = (0/0)? ∴must be differentiable?
$$\mathrm{if}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{b}_{{n}} =\mathrm{0} \\ $$$$\mathrm{Doesn}'\mathrm{t}\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{a}_{{n}} }{{b}_{{n}} }\:=\:\frac{\mathrm{0}}{\mathrm{0}}? \\ $$$$ \\ $$$$\therefore{must}\:{be}\:{differentiable}? \\ $$
Commented by Filup last updated on 14/Dec/15
sorry, i missread b_n ^(1/n) as b_n ^(−1) and i was reffering to how if x_n converges, lim_(n→∞) x_n =0. if i am not mistaken
$${sorry},\:{i}\:{missread}\:{b}_{{n}} ^{\mathrm{1}/{n}} \:{as}\:{b}_{{n}} ^{−\mathrm{1}} \\ $$$$ \\ $$$${and}\:{i}\:{was}\:{reffering}\:{to}\:{how}\:{if} \\ $$$${x}_{{n}} \:{converges},\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{x}_{{n}} =\mathrm{0}.\:\mathrm{if}\:\mathrm{i}\:\mathrm{am}\:\mathrm{not}\:\mathrm{mistaken} \\ $$
Commented by 123456 last updated on 14/Dec/15
if im not wrong. we asking for sulficient conditions, lim_(n→∞) a_n =0 is only a necessary, but not sulficient (a_n =1/n is any example a_n →0 as n→∞, but Σa_n diverged)
$$\mathrm{if}\:\mathrm{im}\:\mathrm{not}\:\mathrm{wrong}. \\ $$$$\mathrm{we}\:\mathrm{asking}\:\mathrm{for}\:\mathrm{sulficient}\:\mathrm{conditions}, \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} =\mathrm{0}\:\mathrm{is}\:\mathrm{only}\:\mathrm{a}\:\mathrm{necessary},\:\mathrm{but}\:\mathrm{not} \\ $$$$\mathrm{sulficient}\:\left({a}_{{n}} =\mathrm{1}/{n}\:\mathrm{is}\:\mathrm{any}\:\mathrm{example}\right. \\ $$$$\left.{a}_{{n}} \rightarrow\mathrm{0}\:\mathrm{as}\:{n}\rightarrow\infty,\:\mathrm{but}\:\Sigma{a}_{{n}} \:\mathrm{diverged}\right) \\ $$
Commented by Yozzii last updated on 14/Dec/15
s=Σ_(n=1) ^∞ a_n ^n b_n ^(1/n) e^s =e^(Σ_(n=1) ^∞ a_n ^n b_n ^(1/n) ) e^s =e^(a_1 b_1 ) e^(a_2 ^2 b_2 ^(1/2) ) e^(a_3 ^3 b_3 ^(1/3) ) e^(a_4 ^4 b_4 ^(1/4) ) .... e^s =Π_(n=1) ^∞ {e^(a_n ^n b_n ^(1/n) ) −1+1} let g_n =e^(a_n ^n b_n ^(1/n) ) −1 ∴ e^s =Π_(n=1) ^∞ (g_n +1) if e^s is defined, the infinite product on the rhs converges. If this is so, then Σ_(n=1) ^∞ g_n converges. Σ_(n=1) ^∞ (e^(a_n ^n b_n ^(1/n) ) −1)=Σ_(n=1) ^∞ e^(a_n ^n b_n ^(1/n) ) −∞ If lim_(n→∞) g_n =0 lim_(n→∞) e^(a_n ^n b_n ^(1/n) ) =1
$${s}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} \\ $$$${e}^{{s}} ={e}^{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} } \\ $$$${e}^{{s}} ={e}^{{a}_{\mathrm{1}} {b}_{\mathrm{1}} } {e}^{{a}_{\mathrm{2}} ^{\mathrm{2}} {b}_{\mathrm{2}} ^{\mathrm{1}/\mathrm{2}} } {e}^{{a}_{\mathrm{3}} ^{\mathrm{3}} {b}_{\mathrm{3}} ^{\mathrm{1}/\mathrm{3}} } {e}^{{a}_{\mathrm{4}} ^{\mathrm{4}} {b}_{\mathrm{4}} ^{\mathrm{1}/\mathrm{4}} } …. \\ $$$${e}^{{s}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left\{{e}^{{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} } −\mathrm{1}+\mathrm{1}\right\} \\ $$$${let}\:{g}_{{n}} ={e}^{{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} } −\mathrm{1} \\ $$$$\therefore\:{e}^{{s}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left({g}_{{n}} +\mathrm{1}\right) \\ $$$${if}\:{e}^{{s}} \:{is}\:{defined},\:{the}\:{infinite}\:{product} \\ $$$${on}\:{the}\:{rhs}\:{converges}.\:{If}\:{this}\:{is}\:{so}, \\ $$$${then}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{g}_{{n}} \:{converges}.\: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({e}^{{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} } −\mathrm{1}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{e}^{{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} } −\infty\: \\ $$$${If}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{g}_{{n}} =\mathrm{0} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{e}^{{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} } =\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by prakash jain last updated on 14/Dec/15
Yes. The question is asking for sufficient condition.
$$\mathrm{Yes}.\:\mathrm{The}\:\mathrm{question}\:\mathrm{is}\:\mathrm{asking}\:\mathrm{for}\:\mathrm{sufficient} \\ $$$$\mathrm{condition}. \\ $$
Commented by Yozzii last updated on 14/Dec/15
Yes...
$${Yes}… \\ $$
Commented by prakash jain last updated on 14/Dec/15
lim_(n→0) e^(a_n ^n b_n ^(1/n) ) =1 if lim_(n→0) a_n ^n b_n ^(1/n) =0 Since c_n =a_n ^n b_n ^(1/n) this is a necessary condition that lim_(n→∞) c_n =0, but this is not sufficient.
$$\underset{{n}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} } =\mathrm{1}\:{if} \\ $$$$\underset{{n}\rightarrow\mathrm{0}} {\mathrm{lim}}{a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} =\mathrm{0} \\ $$$$\mathrm{Since}\:{c}_{{n}} ={a}_{{n}} ^{{n}} {b}_{{n}} ^{\mathrm{1}/{n}} \:\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{necessary}\:\mathrm{condition} \\ $$$$\mathrm{that}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{c}_{{n}} =\mathrm{0},\:\mathrm{but}\:\mathrm{this}\:\mathrm{is}\:\mathrm{not}\:\mathrm{sufficient}. \\ $$

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