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Question Number 52214 by Tawa1 last updated on 04/Jan/19

Commented by Tawa1 last updated on 04/Jan/19

$$\mathrm{Evaluate}$$

Commented by mr W last updated on 05/Jan/19

$$iknow{it}^{\prime }s\frac{9}{32}$$$$buti{can}^{\prime }tprove.$$

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

$$\mathrm{Please}\mathrm{show}\mathrm{it}\mathrm{sir}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Jan/19

$$T_{m,n}=\frac{\frac{m}{3^m}}{(\frac{3^m}{m}+\frac{3^n}{n})}$$$$sincebothm\to \mathrm{\infty }$$$$andn\to \mathrm{\infty }$$$$andvalueof\frac{3^m}{m}=\frac{3^n}{n}whenweputm=1,2,3...$$$$n=1,2,3...simutaneously$$$$so$$$$T_{m,n}=T_r=\frac{\frac{r}{3^r}}{2\times \frac{3^r}{r}}=\frac{1}{2}\times {\left(\frac{r}{3^r}\right)}^2$$$$\mathit{s}={\mathit{T}}_1+T_2+T_3+...\mathrm{\infty }$$$$S=\frac{1}{2}[{\left(\frac{1}{3^1}\right)}^2+{\left(\frac{2}{3^2}\right)}^2+{\left(\frac{3}{3^3}\right)}^2+...\mathrm{\infty }$$$$wait...$$

Commented by mr W last updated on 05/Jan/19

$${it}^{\prime }snotcorrectsir.$$$$youcannottransfer{T}_{\mathrm{\infty },\mathrm{\infty }}to{T}_{m,n}.$$$$\underset{m=1}{\sum ^{\mathrm{\infty }}}\underset{n=1}{\sum ^{\mathrm{\infty }}}{a}_{m,n}=(a_{1,1}+a_{1,2}+...)+(a_{2,1}+a_{2,3}+...)+(.....)+$$$$=\left(\mathrm{\infty }numbers\right)+\left(\mathrm{\infty }numbers\right)+(...)+$$$$=\mathrm{\infty }\times \mathrm{\infty }numbers$$$$$$$$e.g.$$$$\underset{m=1}{\sum ^{\mathrm{\infty }}}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{m^2+n^2}\ne \frac{1}{2}\underset{r=1}{\sum ^{\mathrm{\infty }}}\frac{1}{r^2}$$$$$$$$tocheck:$$$$S=\frac{1}{2}[{\left(\frac{1}{3^1}\right)}^2+{\left(\frac{2}{3^2}\right)}^2+{\left(\frac{3}{3^3}\right)}^2+...\mathrm{\infty }=\frac{45}{512}\ne \frac{9}{32}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Jan/19

$$thankyousir...ivieweditdiffdrently...$$

Answered by rahul 19 last updated on 05/Jan/19

$$S=\underset{m=1}{\sum ^{\mathrm{\infty }}}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{m^{2}n}{3^m(n.3^m+m.3^n)}$$$$S=\underset{n=1}{\sum ^{\mathrm{\infty }}}\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{n^{2}m}{3^n(m.3^n+n.3^m)}$$$$\Rightarrow 2S=\mathrm{\Sigma }\mathrm{\Sigma }\frac{mn}{n.3^m+m.3^n}[\frac{m}{3^m}+\frac{n}{3^n}].$$$$\Rightarrow 2S=\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{m}{3^m}.\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n}{3^n}$$$$\Rightarrow 2S=\frac{3}{4}\times \frac{3}{4}$$$$\Rightarrow S=\frac{9}{32}.$$

Commented by mr W last updated on 05/Jan/19

$$verytrickysir!$$

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