# Thread: Convergence in the form p^n/n!

1. ## Convergence in the form p^n/n!

Are these sufficient answers? I feel like maybe they are not convincing enough...

2. Originally Posted by davismj

Are these sufficient answers? I feel like maybe they are not convincing enough...
They don't convince me at all. Can you explain your reasoning?

3. Originally Posted by Drexel28
They don't convince me at all. Can you explain your reasoning?
well, any complex number $\displaystyle z$ can be written $\displaystyle \rho(e^i\theta)$, and $\displaystyle \rho \in R : \rho = |z|$.

Therefore $\displaystyle \sum\rho^n/n! = e^\rho$.

4. Originally Posted by davismj
well, any complex number $\displaystyle z$ can be written $\displaystyle \rho(e^i\theta)$, and $\displaystyle \rho \in R : \rho = |z|$.

Therefore $\displaystyle \sum\rho^n/n! = e^\rho$.
Yes, but $\displaystyle \lim\text{ }a_n=0\implies \sum_{n=1}^{\infty}a_n<\infty$?

5. Originally Posted by Drexel28
Yes, but $\displaystyle \lim\text{ }a_n=0\implies \sum_{n=1}^{\infty}a_n<\infty$?
No, although in this case we know that since $\displaystyle \rho$ is real, the sum is actually equal to $\displaystyle e^\rho$. Should I write that down?

Thank you for your help!

6. Originally Posted by davismj
No, although in this case we know that since $\displaystyle \rho$ is real, the sum is actually equal to $\displaystyle e^\rho$. Should I write that down?

Thank you for your help!
That isn't how to do this. Try using the ratio test.

7. Originally Posted by Drexel28
That isn't how to do this. Try using the ratio test.
I see now. I was making things harder than necessary.