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Question Number 120160 by 77731 last updated on 29/Oct/20

Commented by JDamian last updated on 29/Oct/20

https://m.youtube.com/watch?v=9kfmk7jEuvQ

Answered by mindispower last updated on 29/Oct/20

$$\frac{1}{2^m(2^n+2^m)}=-\frac{1}{2^n(2^n+2^m)}+\frac{1}{2^{n+m}}$$$$\sum _{m⩾0}\sum _{n⩾0}\frac{m+n+nm}{2^m(2^n+2^m)}=-\sum _{m⩾0}\sum _{n⩾0}\frac{m+n+nm}{2^n(2^n+2^m)}+\mathrm{\Sigma }\mathrm{\Sigma }\frac{n+m+nm}{2^{n+m}}$$$$S=\sum _{m⩾0}\sum _{n⩾0}\frac{m+n+nm}{2^m(2^n+2^m)}=\sum _m\sum _n\frac{n+m+nm}{2^n(2^m+2^n)}=S$$$$(n,m)\to (m,n)isbijection$$$$\iff$$$$S=-S+2\sum _m\sum _n\frac{m}{2^{n+m}}+\mathrm{\Sigma }\mathrm{\Sigma }\frac{mn}{2^n.2^m}$$$$\sum _{m⩾0}\frac{m}{2^m}=\sum _{m⩾1}\frac{m}{2^m},\frac{1}{1-x}=\mathrm{\Sigma }{x}^n\Rightarrow \frac{x}{{(1-x)}^2}=\sum _{n⩾1}{nx}^n$$$$2S=2.\frac{1}{1-\frac{1}{2}}.\frac{\frac{1}{2}}{{(1-\frac{1}{2})}^2}+{\left(\frac{\frac{1}{2}}{{(1-\frac{1}{2})}^2}\right)}^2$$$$2S=4.2+4\Rightarrow S=6$$$$\sum _{n⩾0}\sum _{m⩾0}\frac{n+m+nm}{2^m(2^m+2^n)}=6$$

Commented by 77731 last updated on 01/Nov/20

very nice work ...!!

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