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Question-52214




Question Number 52214 by Tawa1 last updated on 04/Jan/19
Commented by Tawa1 last updated on 04/Jan/19
Evaluate
$$\mathrm{Evaluate} \\ $$
Commented by mr W last updated on 05/Jan/19
i know it′s (9/(32))  but i can′t prove.
$${i}\:{know}\:{it}'{s}\:\frac{\mathrm{9}}{\mathrm{32}} \\ $$$${but}\:{i}\:{can}'{t}\:{prove}. \\ $$
Commented by rahul 19 last updated on 05/Jan/19
Yes Sir! U r right . Just interchange the summation (as they are independent) , call them as EQ. (1) & (2) and add them...��
Commented by Tawa1 last updated on 05/Jan/19
Please show it sir
$$\mathrm{Please}\:\mathrm{show}\:\mathrm{it}\:\mathrm{sir} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 05/Jan/19
T_(m,n) =((m/3^m )/(((3^m /m)+(3^n /n))))  since both m→∞  and n→∞  and  value of (3^m /m)=(3^n /n) when we put m=1,2,3...  n=1,2,3...simutaneously  so   T_(m,n) =T_(r ) =((r/3^r )/(2×(3^r /r)))=(1/2)×((r/3^r ))^2   s=T_1 +T_2 +T_3 +...∞  S=(1/2)[((1/3^1 ))^2 +((2/3^2 ))^2 +((3/3^3 ))^2 +...∞  wait...
$${T}_{{m},{n}} =\frac{\frac{{m}}{\mathrm{3}^{{m}} }}{\left(\frac{\mathrm{3}^{{m}} }{{m}}+\frac{\mathrm{3}^{{n}} }{{n}}\right)} \\ $$$${since}\:{both}\:{m}\rightarrow\infty \\ $$$${and}\:{n}\rightarrow\infty \\ $$$${and}\:\:{value}\:{of}\:\frac{\mathrm{3}^{{m}} }{{m}}=\frac{\mathrm{3}^{{n}} }{{n}}\:{when}\:{we}\:{put}\:{m}=\mathrm{1},\mathrm{2},\mathrm{3}… \\ $$$${n}=\mathrm{1},\mathrm{2},\mathrm{3}…{simutaneously} \\ $$$${so}\: \\ $$$${T}_{{m},{n}} ={T}_{{r}\:} =\frac{\frac{{r}}{\mathrm{3}^{{r}} }}{\mathrm{2}×\frac{\mathrm{3}^{{r}} }{{r}}}=\frac{\mathrm{1}}{\mathrm{2}}×\left(\frac{{r}}{\mathrm{3}^{{r}} }\right)^{\mathrm{2}} \\ $$$$\boldsymbol{{s}}=\boldsymbol{{T}}_{\mathrm{1}} +{T}_{\mathrm{2}} +{T}_{\mathrm{3}} +…\infty \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{1}} }\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}} }\right)^{\mathrm{2}} +\left(\frac{\mathrm{3}}{\mathrm{3}^{\mathrm{3}} }\right)^{\mathrm{2}} +…\infty\right. \\ $$$${wait}… \\ $$
Commented by mr W last updated on 05/Jan/19
it′s not correct sir.  you can not transfer T_(∞,∞)  to T_(m,n) .  Σ_(m=1) ^∞  Σ_(n=1) ^∞ a_(m,n) =(a_(1,1) +a_(1,2) +...)+(a_(2,1) +a_(2,3) +...)+(.....)+  =(∞ numbers)+(∞ numbers)+(...)+  =∞×∞ numbers    e.g.  Σ_(m=1) ^∞  Σ_(n=1) ^∞ (1/(m^2 +n^2 ))≠(1/2)Σ_(r=1) ^∞ (1/r^2 )    to check:  S=(1/2)[((1/3^1 ))^2 +((2/3^2 ))^2 +((3/3^3 ))^2 +...∞=((45)/(512))≠(9/(32))
$${it}'{s}\:{not}\:{correct}\:{sir}. \\ $$$${you}\:{can}\:{not}\:{transfer}\:{T}_{\infty,\infty} \:{to}\:{T}_{{m},{n}} . \\ $$$$\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{m},{n}} =\left({a}_{\mathrm{1},\mathrm{1}} +{a}_{\mathrm{1},\mathrm{2}} +…\right)+\left({a}_{\mathrm{2},\mathrm{1}} +{a}_{\mathrm{2},\mathrm{3}} +…\right)+\left(…..\right)+ \\ $$$$=\left(\infty\:{numbers}\right)+\left(\infty\:{numbers}\right)+\left(…\right)+ \\ $$$$=\infty×\infty\:{numbers} \\ $$$$ \\ $$$${e}.{g}. \\ $$$$\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{m}^{\mathrm{2}} +{n}^{\mathrm{2}} }\neq\frac{\mathrm{1}}{\mathrm{2}}\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{r}^{\mathrm{2}} } \\ $$$$ \\ $$$${to}\:{check}: \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{1}} }\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}} }\right)^{\mathrm{2}} +\left(\frac{\mathrm{3}}{\mathrm{3}^{\mathrm{3}} }\right)^{\mathrm{2}} +…\infty=\frac{\mathrm{45}}{\mathrm{512}}\neq\frac{\mathrm{9}}{\mathrm{32}}\right. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Jan/19
thank you sir...i viewed it diffdrently...
$${thank}\:{you}\:{sir}…{i}\:{viewed}\:{it}\:{diffdrently}… \\ $$
Answered by rahul 19 last updated on 05/Jan/19
S = Σ_(m=1) ^∞  Σ_(n=1) ^∞  ((m^2 n)/(3^m (n.3^m +m.3^n )))  S =  Σ_(n=1) ^∞ Σ_(m=1) ^∞ ((n^2 m)/(3^n (m.3^n +n.3^m )))  ⇒ 2S = ΣΣ((mn)/(n.3^m +m.3^n ))[(m/3^m )+(n/3^n )].  ⇒2S= Σ_(m=1) ^∞ (m/3^m )  . Σ_(n=1) ^∞ (n/3^n )  ⇒2S= (3/4) × (3/4)  ⇒ S = (9/(32 )) .
$${S}\:=\:\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{m}^{\mathrm{2}} {n}}{\mathrm{3}^{{m}} \left({n}.\mathrm{3}^{{m}} +{m}.\mathrm{3}^{{n}} \right)} \\ $$$${S}\:=\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} {m}}{\mathrm{3}^{{n}} \left({m}.\mathrm{3}^{{n}} +{n}.\mathrm{3}^{{m}} \right)} \\ $$$$\Rightarrow\:\mathrm{2}{S}\:=\:\Sigma\Sigma\frac{{mn}}{{n}.\mathrm{3}^{{m}} +{m}.\mathrm{3}^{{n}} }\left[\frac{{m}}{\mathrm{3}^{{m}} }+\frac{{n}}{\mathrm{3}^{{n}} }\right]. \\ $$$$\Rightarrow\mathrm{2}{S}=\:\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{m}}{\mathrm{3}^{{m}} }\:\:.\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{\mathrm{3}^{{n}} } \\ $$$$\Rightarrow\mathrm{2}{S}=\:\frac{\mathrm{3}}{\mathrm{4}}\:×\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\:{S}\:=\:\frac{\mathrm{9}}{\mathrm{32}\:}\:. \\ $$
Commented by mr W last updated on 05/Jan/19
very tricky sir!
$${very}\:{tricky}\:{sir}! \\ $$

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