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Question Number 3716 by Yozzii last updated on 19/Dec/15
Σ_(m=0) ^∞ Σ_(n=0) ^∞ ((m+n+mn)/(2^m (2^m +2^n )))=? (Not originally produced by me.)
$$\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{m}+{n}+{mn}}{\mathrm{2}^{{m}} \left(\mathrm{2}^{{m}} +\mathrm{2}^{{n}} \right)}=?\: \\ $$$$\left({Not}\:{originally}\:{produced}\:{by}\:{me}.\right) \\ $$
Commented by prakash jain last updated on 19/Dec/15
Σ_(n=0) ^∞ (n/2^n )=2 Σ_(m=0) ^∞ Σ_(n=0) ^∞ ((nm)/2^(m+n) )=4 Σ_(m=0) ^∞ Σ_(n=0) ^∞ (n/2^(m+n) ) =Σ_(m=0) ^∞ (1/2^m ) Σ_(n=0) ^∞ (n/2^n )=Σ_(m=0) ^∞ (2/2^m ) =4 Σ_(m=0) ^∞ Σ_(n=0) ^∞ (m/2^(m+n) ) = Σ_(m=0) ^∞ (m/2^m ) Σ_(n=0) ^∞ (1/2^n )= Σ_(m=0) ^∞ ((2m)/2^m ) =4 Σ_(m=0) ^∞ Σ_(n=0) ^∞ ((m+n+nm)/2^(m+n) )=12
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{n}}{\mathrm{2}^{{n}} }=\mathrm{2} \\ $$$$\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{nm}}{\mathrm{2}^{{m}+{n}} }=\mathrm{4} \\ $$$$\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{n}}{\mathrm{2}^{{m}+{n}} }\:=\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{m}} }\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{n}}{\mathrm{2}^{{n}} }=\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\mathrm{2}^{{m}} }\:=\mathrm{4} \\ $$$$\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{m}}{\mathrm{2}^{{m}+{n}} }\:=\:\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{m}}{\mathrm{2}^{{m}} }\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }=\:\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{m}}{\mathrm{2}^{{m}} }\:=\mathrm{4} \\ $$$$\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{m}+{n}+{nm}}{\mathrm{2}^{{m}+{n}} }=\mathrm{12} \\ $$$$ \\ $$
Commented by prakash jain last updated on 19/Dec/15
case 1: m=n=0 ((m+n+mn)/(2^m (2^m +2^n )))=0 ((m+n+nm)/2^(m+n) )=0 case 2:m and are both not 0 ((m+n+mn)/(2^m (2^m +2^n )))−((m+n+nm)/2^(m+n) ) (m+n+mn)[(1/(2^m (2^m +2^n )))−(1/2^(m+n) )] ((m+n+nm)/(2^m 2^(m+n) (2^m +2^n )))(2^(m+n) −2^(2m) −2^(m+n) ) =−((2^(2m) (m+n+mn))/(2^m 2^(m+n) (2^m +2^n )))<0 for all m,n so Σ_(m=0) ^∞ Σ_(n=0) ^∞ ((m+n+mn)/(2^m (2^m +2^n ))) converges.
$${case}\:\mathrm{1}:\:{m}={n}=\mathrm{0} \\ $$$$\frac{{m}+{n}+{mn}}{\mathrm{2}^{{m}} \left(\mathrm{2}^{{m}} +\mathrm{2}^{{n}} \right)}=\mathrm{0}\:\:\frac{{m}+{n}+{nm}}{\mathrm{2}^{{m}+{n}} }=\mathrm{0} \\ $$$$\mathrm{case}\:\mathrm{2}:{m}\:{and}\:{are}\:{both}\:{not}\:\mathrm{0} \\ $$$$\frac{{m}+{n}+{mn}}{\mathrm{2}^{{m}} \left(\mathrm{2}^{{m}} +\mathrm{2}^{{n}} \right)}−\frac{{m}+{n}+{nm}}{\mathrm{2}^{{m}+{n}} } \\ $$$$\left({m}+{n}+{mn}\right)\left[\frac{\mathrm{1}}{\mathrm{2}^{{m}} \left(\mathrm{2}^{{m}} +\mathrm{2}^{{n}} \right)}−\frac{\mathrm{1}}{\mathrm{2}^{{m}+{n}} }\right] \\ $$$$\frac{{m}+{n}+{nm}}{\mathrm{2}^{{m}} \mathrm{2}^{{m}+{n}} \left(\mathrm{2}^{{m}} +\mathrm{2}^{{n}} \right)}\left(\mathrm{2}^{{m}+{n}} −\mathrm{2}^{\mathrm{2}{m}} −\mathrm{2}^{{m}+{n}} \right) \\ $$$$=−\frac{\mathrm{2}^{\mathrm{2}{m}} \left({m}+{n}+{mn}\right)}{\mathrm{2}^{{m}} \mathrm{2}^{{m}+{n}} \left(\mathrm{2}^{{m}} +\mathrm{2}^{{n}} \right)}<\mathrm{0}\:{for}\:{all}\:{m},{n} \\ $$$${so} \\ $$$$\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{m}+{n}+{mn}}{\mathrm{2}^{{m}} \left(\mathrm{2}^{{m}} +\mathrm{2}^{{n}} \right)}\:\mathrm{converges}. \\ $$
Commented by Yozzii last updated on 19/Dec/15
How is Σ_(m=0) ^∞ Σ_(n=1) ^∞ ((m+n+mn)/2^(m+n) )=12 ? I′m wondering how you evaluated the double summation. I understood the two first results.
$${How}\:{is}\:\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{m}+{n}+{mn}}{\mathrm{2}^{{m}+{n}} }=\mathrm{12}\:? \\ $$$${I}'{m}\:{wondering}\:{how}\:{you}\:{evaluated}\:{the} \\ $$$${double}\:{summation}.\:{I}\:{understood}\:{the} \\ $$$${two}\:{first}\:{results}.\: \\ $$
Commented by prakash jain last updated on 19/Dec/15
Updated comments for the results.
$$\mathrm{Updated}\:\mathrm{comments}\:\mathrm{for}\:\mathrm{the}\:\mathrm{results}. \\ $$
Commented by Yozzii last updated on 19/Dec/15
Thanks.
$${Thanks}. \\ $$
Answered by prakash jain last updated on 19/Dec/15
Not sure of the answer. Please verify ((m+n+nm)/(2^m (2^m +2^n )))=((m+n+nm)/2^n )[(1/2^m )−(1/(2^n +2^m ))] Σ_(m=0) ^∞ Σ_(n=0) ^∞ ((m+n+nm)/(2^m (2^m +2^n ))) =Σ_(m=0) ^∞ Σ_(n=0) ^∞ ((m+n+nm)/2^(n+m) )−Σ_(m=0) ^∞ Σ_(n=0) ^∞ ((m+n+nm)/(2^n (2^m +2^n ))) Σ_(m=0) ^∞ Σ_(n=0) ^∞ ((m+n+nm)/(2^m (2^m +2^n )))=S due to symmetry Σ_(m=0) ^∞ Σ_(n=0) ^∞ ((m+n+nm)/(2^n (2^m +2^n )))=S S=Σ_(m=0) ^∞ Σ_(n=0) ^∞ ((m+n+nm)/2^(n+m) )−S 2S=Σ_(m=0) ^∞ Σ_(n=0) ^∞ ((m+n+nm)/2^(n+m) )=12 S=6
$$\mathrm{Not}\:\mathrm{sure}\:\mathrm{of}\:\mathrm{the}\:\mathrm{answer}.\:\mathrm{Please}\:\mathrm{verify} \\ $$$$\frac{{m}+{n}+{nm}}{\mathrm{2}^{{m}} \left(\mathrm{2}^{{m}} +\mathrm{2}^{{n}} \right)}=\frac{{m}+{n}+{nm}}{\mathrm{2}^{{n}} }\left[\frac{\mathrm{1}}{\mathrm{2}^{{m}} }−\frac{\mathrm{1}}{\mathrm{2}^{{n}} +\mathrm{2}^{{m}} }\right] \\ $$$$\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{m}+{n}+{nm}}{\mathrm{2}^{{m}} \left(\mathrm{2}^{{m}} +\mathrm{2}^{{n}} \right)} \\ $$$$\:\:\:\:\:\:=\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{m}+{n}+{nm}}{\mathrm{2}^{{n}+{m}} }−\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{m}+{n}+{nm}}{\mathrm{2}^{{n}} \left(\mathrm{2}^{{m}} +\mathrm{2}^{{n}} \right)} \\ $$$$\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{m}+{n}+{nm}}{\mathrm{2}^{{m}} \left(\mathrm{2}^{{m}} +\mathrm{2}^{{n}} \right)}=\mathrm{S} \\ $$$$\mathrm{due}\:\mathrm{to}\:\mathrm{symmetry}\:\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{m}+{n}+{nm}}{\mathrm{2}^{{n}} \left(\mathrm{2}^{{m}} +\mathrm{2}^{{n}} \right)}=\mathrm{S} \\ $$$$\mathrm{S}=\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{m}+{n}+{nm}}{\mathrm{2}^{{n}+{m}} }−\mathrm{S} \\ $$$$\mathrm{2S}=\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{m}+{n}+{nm}}{\mathrm{2}^{{n}+{m}} }=\mathrm{12} \\ $$$$\mathrm{S}=\mathrm{6} \\ $$
Commented by Yozzii last updated on 20/Dec/15
Looks correct. Thanks!
$${Looks}\:{correct}.\:{Thanks}! \\ $$

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