An expression resembling Laguerre

  • #1
sarrah1
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This was posted to calculus forum. I suppose it should have been posted here.

I am trying to find a closed form expression/or limit as $n\implies\infty$ of

${S}_{n}=\sum_{k=0}^{n}{n \choose k} {a}^{k} \sum_{j=0}^{n}{n \choose j}\frac{{b}^{n-j}{c}^{j}}{(k+j)!}$

where $a$ , $b$ and $c$ are positive constants

the ultimate aim is to find the limit of ${S}_{n}$ as $n$ tends to infinity and that this limit is zero under some conditions imposed upon the constants a,b,c.

Euge gave me once the idea for a similar expression by taking $k+j$ outside the 2nd sum and the remaining becomes equal to ${(b+c)}^{n}$ . As for the first term, when divided by $k!$ it will be less or equal to

$\sum_{k=0}^{n}{n \choose k} {a}^{k}/k!$

So if this one converges to some sum that tends to zero, OR Instead to a finite value, But $b+c<1$ then mission done and ${S}_{n}$ tends to zero.

On a second thought the expression $\sum_{k=0}^{n}{n \choose k} {a}^{k}/k!$ is a Laguerre polynomial ${L}_{n}(-a) , a>0 $ which diverges $n\implies\infty$ . so one has to rely on the whole expression ${S}_{n}$ above. If someone knows if there is some double Laguerre, or product of 2 laguerre polynomials or double binomial transform, etc...
I shall be grateful
happy new prosperous year
special regards to Euge, Oplag and Akbach

sarrah
 
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  • #2
Dear Oplag
I received this in my mailbox:

I should have looked a bit further before posting that. Apparently it is my browser (Safari) that is at fault, because when I look at the same pages with Firefox, the formulas are correctly rendered.
***************There may also be other replies, but you will not receive any more notifications until you visit the forum again.

All the best,
Math Help Boards | Free Math Help

I couldn't track your letter, unless I didn't get what you meant
sarrah
 
  • #3
sarrah said:
This was posted to calculus forum. I suppose it should have been posted here.

I am trying to find a closed form expression/or limit as $n\implies\infty$ of

${S}_{n}=\sum_{k=0}^{n}{n \choose k} {a}^{k} \sum_{j=0}^{n}{n \choose j}\frac{{b}^{n-j}{c}^{j}}{(k+j)!}$

where $a$ , $b$ and $c$ are positive constants

the ultimate aim is to find the limit of ${S}_{n}$ as $n$ tends to infinity and that this limit is zero under some conditions imposed upon the constants a,b,c.

Euge gave me once the idea for a similar expression by taking $k+j$ outside the 2nd sum and the remaining becomes equal to ${(b+c)}^{n}$ . As for the first term, when divided by $k!$ it will be less or equal to

$\sum_{k=0}^{n}{n \choose k} {a}^{k}/k!$

So if this one converges to some sum that tends to zero, OR Instead to a finite value, But $b+c<1$ then mission done and ${S}_{n}$ tends to zero.

On a second thought the expression $\sum_{k=0}^{n}{n \choose k} {a}^{k}/k!$ is a Laguerre polynomial ${L}_{n}(-a) , a>0 $ which diverges $n\implies\infty$ . so one has to rely on the whole expression ${S}_{n}$ above. If someone knows if there is some double Laguerre, or product of 2 laguerre polynomials or double binomial transform, etc...
I shall be grateful
happy new prosperous year
special regards to Euge, Oplag and Akbach

sarrah

Dear all
incidentally, i gave values for $a,b,c$ like equal to 1, it was divergent. When I gave the value 0.5 to all, it converged. So there must be a condition on the constants to promote convergence
sarrah
 

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