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Question Number 192590 by York12 last updated on 22/May/23
Σ_(n=1) ^∞ Σ_(m=1) ^∞ [((m^2 n)/(3^m (3^n .m+3^m .n)))]=λ find λ.
$$ \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{{m}^{\mathrm{2}} {n}}{\mathrm{3}^{{m}} \left(\mathrm{3}^{{n}} .{m}+\mathrm{3}^{{m}} .{n}\right)}\right]=\lambda\: \\ $$$${find}\:\lambda. \\ $$$$ \\ $$
Answered by witcher3 last updated on 24/May/23
(n,m)→^f ((m^2 n)/(3^m (3^n m+3^m n))) ∀n∈Z_+ ,Σ_(m≥1) f(n,m)=nΣ_(m≥1) (m^2 /(3^m (3^n m+3^m n))) ≤nΣ_(m≥1) (m^2 /3^m )=S ∃N∈N ∀m≥N,m^2 ≤2^m S≤Σ_(m=1) ^(N−1) n(m^2 /3^m )+nΣ_N ((2/3))^m <∞ ∀m∈Z_+ f(n,m)=(m^2 /3^m )Σ(n/((3^n m+3^m n)))≤(m^2 /3^m )Σ(n/3^n ) Sam as previous <∞ ⇒Σ_n Σ_m =Σ_m Σ_n ΣΣf(n,m)=ΣΣf(m,n)=S 2S=Σ_n Σ_m ((m^2 n)/(3^m (n3^m +m3^n )))+((n^2 m)/(3^n (n3^m +m3^n ))) =Σ_n Σ_m ((mn(m3^n +n3^m ))/(3^(m+n) (n3^m +m3^n )))=Σ_n (n/3^n )Σ(m/3^m ) (1/((1−x)^2 ))=Σ_(n≥1) nx^(n−1) ⇒(x/((1−x)^2 ))=Σnx^n S=(((1/3)/((1−(1/3))^2 )))^2 =(9/(16))
$$\left(\mathrm{n},\mathrm{m}\right)\overset{\mathrm{f}} {\rightarrow}\frac{\mathrm{m}^{\mathrm{2}} \mathrm{n}}{\mathrm{3}^{\mathrm{m}} \left(\mathrm{3}^{\mathrm{n}} \mathrm{m}+\mathrm{3}^{\mathrm{m}} \mathrm{n}\right)} \\ $$$$\forall\mathrm{n}\in\mathbb{Z}_{+} ,\underset{\mathrm{m}\geqslant\mathrm{1}} {\sum}\mathrm{f}\left(\mathrm{n},\mathrm{m}\right)=\mathrm{n}\underset{\mathrm{m}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{m}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{m}} \left(\mathrm{3}^{\mathrm{n}} \mathrm{m}+\mathrm{3}^{\mathrm{m}} \mathrm{n}\right)} \\ $$$$\leqslant\mathrm{n}\underset{\mathrm{m}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{m}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{m}} }=\mathrm{S} \\ $$$$\exists\mathrm{N}\in\mathbb{N}\:\:\:\forall\mathrm{m}\geqslant\mathrm{N},\mathrm{m}^{\mathrm{2}} \leqslant\mathrm{2}^{\mathrm{m}} \\ $$$$\mathrm{S}\leqslant\underset{\mathrm{m}=\mathrm{1}} {\overset{\mathrm{N}−\mathrm{1}} {\sum}}\mathrm{n}\frac{\mathrm{m}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{m}} }+\mathrm{n}\underset{\mathrm{N}} {\sum}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{m}} <\infty \\ $$$$\forall\mathrm{m}\in\mathbb{Z}_{+} \mathrm{f}\left(\mathrm{n},\mathrm{m}\right)=\frac{\mathrm{m}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{m}} }\Sigma\frac{\mathrm{n}}{\left(\mathrm{3}^{\mathrm{n}} \mathrm{m}+\mathrm{3}^{\mathrm{m}} \mathrm{n}\right)}\leqslant\frac{\mathrm{m}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{m}} }\Sigma\frac{\mathrm{n}}{\mathrm{3}^{\mathrm{n}} } \\ $$$$\mathrm{Sam}\:\mathrm{as}\:\mathrm{previous}\:<\infty \\ $$$$\Rightarrow\underset{\mathrm{n}} {\sum}\underset{\mathrm{m}} {\sum}=\underset{\mathrm{m}} {\sum}\underset{\mathrm{n}} {\sum} \\ $$$$\Sigma\Sigma\mathrm{f}\left(\mathrm{n},\mathrm{m}\right)=\Sigma\Sigma\mathrm{f}\left(\mathrm{m},\mathrm{n}\right)=\mathrm{S} \\ $$$$\mathrm{2S}=\underset{\mathrm{n}} {\sum}\underset{\mathrm{m}} {\sum}\frac{\mathrm{m}^{\mathrm{2}} \mathrm{n}}{\mathrm{3}^{\mathrm{m}} \left(\mathrm{n3}^{\mathrm{m}} +{m}\mathrm{3}^{{n}} \right)}+\frac{{n}^{\mathrm{2}} {m}}{\mathrm{3}^{{n}} \left({n}\mathrm{3}^{{m}} +{m}\mathrm{3}^{{n}} \right)} \\ $$$$=\underset{\mathrm{n}} {\sum}\underset{\mathrm{m}} {\sum}\frac{\mathrm{mn}\left(\mathrm{m3}^{\mathrm{n}} +\mathrm{n3}^{\mathrm{m}} \right)}{\mathrm{3}^{\mathrm{m}+\mathrm{n}} \left(\mathrm{n3}^{\mathrm{m}} +\mathrm{m3}^{\mathrm{n}} \right)}=\underset{\mathrm{n}} {\sum}\frac{\mathrm{n}}{\mathrm{3}^{\mathrm{n}} }\Sigma\frac{\mathrm{m}}{\mathrm{3}^{\mathrm{m}} } \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\mathrm{nx}^{\mathrm{n}−\mathrm{1}} \Rightarrow\frac{\mathrm{x}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }=\Sigma\mathrm{nx}^{\mathrm{n}} \\ $$$$\mathrm{S}=\left(\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} }\right)^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{16}} \\ $$
Commented by York12 last updated on 24/May/23
Great ! what book do you recommend for algebra
$${Great}\:! \\ $$$${what}\:{book}\:{do}\:{you}\:{recommend}\:{for}\:{algebra} \\ $$
Commented by witcher3 last updated on 24/May/23
hello sir sorry for last times not responding i have not telegram book 1 introduce to matrix and linear algebra 2) for calculus almost Impossible sum and integrals contein hards problems !
$$\mathrm{hello}\:\mathrm{sir}\:\mathrm{sorry}\:\mathrm{for}\:\mathrm{last}\:\mathrm{times} \\ $$$$\mathrm{not}\:\mathrm{responding}\:\mathrm{i}\:\mathrm{have}\:\mathrm{not}\:\mathrm{telegram} \\ $$$$\mathrm{book}\:\mathrm{1}\:\mathrm{introduce}\:\mathrm{to}\:\mathrm{matrix}\:\mathrm{and}\:\mathrm{linear}\:\mathrm{algebra} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{for}\:\mathrm{calculus} \\ $$$$\mathrm{almost}\:\mathrm{Impossible}\:\mathrm{sum}\:\mathrm{and}\:\mathrm{integrals} \\ $$$$\mathrm{contein}\:\mathrm{hards}\:\mathrm{problems}\:!\: \\ $$$$ \\ $$
Commented by York12 last updated on 24/May/23
thanks so much , No problem
$${thanks}\:{so}\:{much}\:,\:{No}\:{problem} \\ $$

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