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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 byYozzii 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 byprakash 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 byprakash 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 byFilup 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 byFilup 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 by123456 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 byYozzii 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 byprakash 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 byYozzii last updated on 14/Dec/15

Yes...

$${Yes}... \\ $$

Commented byprakash 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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