- #1
Togli
- 9
- 0
I would like to simplify this series as much as possible
f(m)=\sum_{n=0}^{\infty}\frac{m^n (2n)!}{(n!)^3}
Approximates would also be fine.
One can easily notice that
(2n!) / (n!)^2 > 2^n
hence I figured out that f(m) > \sum_{n=0}^{\infty}\frac{(2m)^n}{n!}=\exp(2m)
but this is not the best approximation, alas, it is far away from the true solution.
What would you suggest? Thanks.
f(m)=\sum_{n=0}^{\infty}\frac{m^n (2n)!}{(n!)^3}
Approximates would also be fine.
One can easily notice that
(2n!) / (n!)^2 > 2^n
hence I figured out that f(m) > \sum_{n=0}^{\infty}\frac{(2m)^n}{n!}=\exp(2m)
but this is not the best approximation, alas, it is far away from the true solution.
What would you suggest? Thanks.
