- #1
Adoniram
- 94
- 6
I am trying to manipulate this summation such that I have a summation of a function of r only by itself somewhere:
[tex]\sum_{r=1}^{\infty}e^{-B⋅r}[/tex]
This could be rewritten:
[tex]\sum_{r=1}^{\infty}\left(e^{-r}\right)^B[/tex] or [tex]\sum_{r=1}^{\infty}\left(e^{-B}\right)^r[/tex]
What I would like is:
[tex]f(r)g(B)[/tex]
or
[tex]g(f(r))[/tex]
Where [itex]g[/itex] depends on [itex]B[/itex] as well.
I really just want to be able to get some kind of independent [itex]f(r)[/itex] out of this. The biggest problem I've had with this is that while the sum converges, doing something like [itex]ln\left[\sum_{r=1}^{\infty}e^{-B⋅r}\right][/itex] does not converge, so I'm stuck...
Thank you!
[tex]\sum_{r=1}^{\infty}e^{-B⋅r}[/tex]
This could be rewritten:
[tex]\sum_{r=1}^{\infty}\left(e^{-r}\right)^B[/tex] or [tex]\sum_{r=1}^{\infty}\left(e^{-B}\right)^r[/tex]
What I would like is:
[tex]f(r)g(B)[/tex]
or
[tex]g(f(r))[/tex]
Where [itex]g[/itex] depends on [itex]B[/itex] as well.
I really just want to be able to get some kind of independent [itex]f(r)[/itex] out of this. The biggest problem I've had with this is that while the sum converges, doing something like [itex]ln\left[\sum_{r=1}^{\infty}e^{-B⋅r}\right][/itex] does not converge, so I'm stuck...
Thank you!
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