Skip to main content

[Community Question] Algebra-Precalculus: series $\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{(m+n-1)!}{m!(n-1)!n!(m-1)!}a^m b^n.$

One of our user asked:

Can anyone please help me with the computation of following series:

$$\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{(m+n-1)!}{m!(n-1)!n!(m-1)!}a^m b^n.$$

My thoughts: Since $$\displaystyle \frac{(m+n-1)!}{m!(n-1)!n!(m-1)!} = \frac{\binom{m+n-1}{m}\binom{m+n-1}{m-1}}{(m+n-1)!},$$ by some arrangement this may be the probability of a hypergeometric distribution.

Any help would be appreciated, thanks!


Labels:

Comments

Post a Comment

Popular posts from this blog

[Community Question] Linear-algebra: Are linear transformations between infinite dimensional vector spaces always differentiable?

One of our user asked: In class we saw that every linear transformation is differentiable (since there's always a linear approximation for them) and we also saw that a differentiable function must be continuous, so it must be true that all linear operators are continuous, however, I just read that between infinite dimensional vector spaces this is not necessarily true. I would like to know where's the flaw in my reasoning (I suspect that linear transformations between infinite dimensional vector spaces are not always differentiable).
Post a Comment
Read more

[Community Question] Calculus: prove $\int_0^\infty \frac{\log^2(x)}{x^2+1}\mathrm dx=\frac{\pi^3}{8}$ with real methods

One of our user asked: I am attempting to prove that $$J=\int_0^\infty\frac{\log^2(x)}{x^2+1}\mathrm dx=\frac{\pi^3}8$$ With real methods because I do not know complex analysis. I have started with the substitution $x=\tan u$ : $$J=\int_0^{\pi/2}\log^2(\tan x)\mathrm dx$$ $$J=\int_0^{\pi/2}\log^2(\cos x)\mathrm dx-2\int_{0}^{\pi/2}\log(\cos x)\log(\sin x)\mathrm dx+\int_0^{\pi/2}\log^2(\sin x)\mathrm dx$$ But frankly, this is basically worse. Could I have some help? Thanks.
Post a Comment
Read more