What Conditions on Constants Make a Series Converge to Zero?

[sarrah1]

Hi

Usually i post in analysis, this time is on combinations so it must be here.

I know that: (although I can't prove it, but I know it's correct)

$\sum_{k=0}^{n}\sum_{j=0}^{m}{n \choose k }{m \choose j} \frac{{x}^{k+j}}{(k+j)!} =\sum_{j=0}^{m+n} {m+n \choose j } \frac{{x}^{j}}{j!}$ ......(1)

in my case I can take $m=n$

But I need the sum of

$\sum_{k=0}^{n}\sum_{j=0}^{n}{n \choose k }{n \choose j} \frac{{a}^{k}{b}^{j}{x}^{k+j}}{(k+j)!} $ where $a,b$ are + constants ......(2)
but in a form like the R.H.S of (1)

my final aim is to find the conditions on $a$ and $b$ such that the sum tends to zero when $n$ tends to $\infty$.
many thanks
sarrah

NB: this is laguerre polynomial which diverges, but possibly the conditions on $a,b$ can make it converge to zero. In fact (1) diverges when $a=b=1$ and converges when $a=b=1/2$

 

[Valkarie]

The solution to the problem posed is as follows:Using the binomial theorem, we can rewrite equation (2) as follows:$\sum_{k=0}^{n}\sum_{j=0}^{n}{n \choose k }{n \choose j} \frac{(a+bx)^k(bx)^j}{(k+j)!}$ By expanding $(a+bx)^k$ and $(bx)^j$, we can see that this expression is equivalent to:$\sum_{j=0}^{n}\sum_{k=0}^{n}{n \choose j }{n \choose k} \frac{a^k b^j x^{k+j}}{(k+j)!}$This expression is identical to the right hand side of equation (1), so it can be rewritten as:$\sum_{j=0}^{2n} {2n \choose j} \frac{x^j}{j!}$Now, in order for the sum to tend to zero when $n$ tends to infinity, we need to choose $a$ and $b$ such that $\frac{a^k b^j x^{k+j}}{(k+j)!}$ goes to zero when $n$ tends to infinity. That is, we need$\lim_{n\to\infty}\frac{a^k b^j x^{k+j}}{(k+j)!} = 0$In other words, we need $a$ and $b$ to be chosen such that $\frac{a^k b^j}{(k+j)!}$ goes to zero as $k+j$ tends to infinity. This can be achieved by setting either $a$ or $b$ to zero. For example, if we set $a=0$, then $\frac{a^k b^j}{(k+j)!}=0$ for all $k>0$. Alternatively, if we set $b=0$, then $\frac{a^k b^j}{(k+j)!}=0$ for all $j>0$. Therefore, the conditions on $a$ and