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Question Number 3735 by prakash jain last updated on 19/Dec/15
Prove  Σ_(m=0) ^∞  Σ_(n=0) ^∞  ((mn)/2^(m+n) )=4
$$\mathrm{Prove} \\ $$$$\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{mn}}{\mathrm{2}^{{m}+{n}} }=\mathrm{4} \\ $$
Commented by prakash jain last updated on 19/Dec/15
As a part answer to Q3716.
$$\mathrm{As}\:\mathrm{a}\:\mathrm{part}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{Q3716}. \\ $$
Answered by prakash jain last updated on 19/Dec/15
Σ_(m=0) ^∞  (m/(2^m  )) Σ_(n=0) ^∞ (n/2^n )=Σ_(m=0) ^∞ ((2m)/2^m )=4  Q3734.
$$\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{m}}{\mathrm{2}^{{m}} \:}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}}{\mathrm{2}^{{n}} }=\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{m}}{\mathrm{2}^{{m}} }=\mathrm{4} \\ $$$${Q}\mathrm{3734}. \\ $$
Commented by Yozzii last updated on 19/Dec/15
What necessary and sufficient   conditions exist so that  Σ_(m∈N) Σ_(n∈N) f(m)g(n)={Σ_(m∈N) f(m)}{Σ_(n∈N) f(n)}   with f,g>0 ?  S=Σ_(m∈N) Σ_(n∈N) f(m)g(m)=Σ_(m∈N) {Σ_(n∈N) f(m)g(n)}  Given that m and are independent  variables, for each m we have   Σ_(n∈N) f(m)g(n)=f(m)×Σ_(n∈N) g(n)  ∴ S=Σ_(m∈N) {f(m)×Σ_(n∈N) g(n)}  If Σ_(n∈N) g(n) converges to l  ⇒S=Σ_(m∈N) f(m)×l=lΣ_(m∈N) f(m)  So, S={Σ_(m∈N) f(m)}×{Σ_(n∈N) g(n)} ?
$${What}\:{necessary}\:{and}\:{sufficient}\: \\ $$$${conditions}\:{exist}\:{so}\:{that} \\ $$$$\underset{{m}\in\mathbb{N}} {\sum}\underset{{n}\in\mathbb{N}} {\sum}{f}\left({m}\right){g}\left({n}\right)=\left\{\underset{{m}\in\mathbb{N}} {\sum}{f}\left({m}\right)\right\}\left\{\underset{{n}\in\mathbb{N}} {\sum}{f}\left({n}\right)\right\}\: \\ $$$${with}\:{f},{g}>\mathrm{0}\:? \\ $$$${S}=\underset{{m}\in\mathbb{N}} {\sum}\underset{{n}\in\mathbb{N}} {\sum}{f}\left({m}\right){g}\left({m}\right)=\underset{{m}\in\mathbb{N}} {\sum}\left\{\underset{{n}\in\mathbb{N}} {\sum}{f}\left({m}\right){g}\left({n}\right)\right\} \\ $$$${Given}\:{that}\:{m}\:{and}\:{are}\:{independent} \\ $$$${variables},\:{for}\:{each}\:{m}\:{we}\:{have}\: \\ $$$$\underset{{n}\in\mathbb{N}} {\sum}{f}\left({m}\right){g}\left({n}\right)={f}\left({m}\right)×\underset{{n}\in\mathbb{N}} {\sum}{g}\left({n}\right) \\ $$$$\therefore\:{S}=\underset{{m}\in\mathbb{N}} {\sum}\left\{{f}\left({m}\right)×\underset{{n}\in\mathbb{N}} {\sum}{g}\left({n}\right)\right\} \\ $$$${If}\:\underset{{n}\in\mathbb{N}} {\sum}{g}\left({n}\right)\:{converges}\:{to}\:{l} \\ $$$$\Rightarrow{S}=\underset{{m}\in\mathbb{N}} {\sum}{f}\left({m}\right)×{l}={l}\underset{{m}\in\mathbb{N}} {\sum}{f}\left({m}\right) \\ $$$${So},\:{S}=\left\{\underset{{m}\in\mathbb{N}} {\sum}{f}\left({m}\right)\right\}×\left\{\underset{{n}\in\mathbb{N}} {\sum}{g}\left({n}\right)\right\}\:? \\ $$$$ \\ $$
Commented by prakash jain last updated on 19/Dec/15
f(m) is constant for summation over n.  taking a constant out for summation is  always valid (even if the series is divergent).  absolute convergence is required for rearrangmnt  changing order of summation.
$${f}\left({m}\right)\:{is}\:{constant}\:{for}\:{summation}\:{over}\:{n}. \\ $$$${taking}\:{a}\:{constant}\:{out}\:{for}\:{summation}\:{is} \\ $$$${always}\:{valid}\:\left({even}\:{if}\:{the}\:{series}\:{is}\:{divergent}\right). \\ $$$${absolute}\:{convergence}\:{is}\:{required}\:{for}\:{rearrangmnt} \\ $$$${changing}\:{order}\:{of}\:{summation}. \\ $$
Commented by Yozzii last updated on 19/Dec/15
Thank you.
$${Thank}\:{you}. \\ $$

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