Sum of series, x >= 0, [ x (n*theta)^x ] / [ (e^{x/n} - 1) x! ]

[Marina5]

Hi!
Could you help me, please, calculate

\(\displaystyle \displaystyle\sum_{x \geq 0} \frac{x\left(n\theta\right)^x}{\left(e^{\frac{x}{n}}-1\right)x!}\)

I know it converges.

Thank you!

 

[mmm4444bot]

Hello Marina. Do you have any beginning work to share? Where in this exercise do you need help?

Here's a link to a summary of the forum's posting guidelines. Thank you! ?
[imath]\;[/imath]

 

[blamocur]

Hi!
Could you help me, please, calculate
\(\displaystyle \displaystyle\sum_{x \geq 0} \frac{x\left(n\theta\right)^x}{\left(e^{\frac{x}{n}}-1\right)x!}\)
I know it converges.
Thank you!

Are you summing by [imath]x[/imath] or by [imath]n[/imath] ?

 

[Marina5]

Are you summing by [imath]x[/imath] or by [imath]n[/imath] ?

by x, n and theta are constants

 

[blamocur]

by x, n and theta are constants

You mean [imath]\sum_{x=0}^\infty ...[/imath] with integer [imath]x[/imath]? Is [imath]n[/imath] an integer or an arbitrary real number ?
Also, what have you tried and which problem have you encountered?

 

[BigBeachBanana]

Looks like the expectation of some conditional Posterior Distribution to find the marginal probability
[math]\Pr(X=x) = \sum x \cdot \Pr(X=x|\theta,n)[/math]