1. ## radius of complex convergence

I stumbled at the answer of a problem solved in Reily-Hobson-Bence.Please check if I am wrong:

Find the parts of the z-plane for which the following series is convergent:

∑[(1/n!)(z^n)] where n runs from 0 to ∞

Changing variables as 1/n=t,the limit becomes (1/R)=Lt(t->0) [(t!)^t]
I am having the limit as 1 and hence,R=1...but the book says Since
[(n!)^(1/n)] behaves like n as n → ∞ we find lim[(1/n!)^(1/n)] = 0. Hence R = ∞ and the series is convergent for all z

2. Originally Posted by kolahalb
I stumbled at the answer of a problem solved in Reily-Hobson-Bence.Please check if I am wrong:

Find the parts of the z-plane for which the following series is convergent:

∑[(1/n!)(z^n)] where n runs from 0 to ∞

Changing variables as 1/n=t,the limit becomes (1/R)=Lt(t->0) [(t!)^t]
I am having the limit as 1 and hence,R=1...but the book says Since
[(n!)^(1/n)] behaves like n as n → ∞ we find lim[(1/n!)^(1/n)] = 0. Hence R = ∞ and the series is convergent for all z
$\frac{1}{n!} \ne (1/n)!$ whatever you think $(1/n)!$ might mean for $n \in \mathbb{N}$.

RonL

3. Yes,I see that. But how to see that [(n!)^(1/n)] behaves like n as n → ∞

4. Originally Posted by kolahalb
Yes,I see that. But how to see that [(n!)^(1/n)] behaves like n as n → ∞
Replace $n!$ by Stirlings approximation and it will appear.

$
n! \sim \sqrt{2 \pi n} n^n e^{-n}
$

RonL

5. Thank you for your help.

I resolved the problem in this manner: (1/n!) falls at a much faster rate than that of (1/n).As a result the function is decreasing all the way starting from the highest value one and always being positive.If you draw a graph,the curve should tend to zero as n tends to infinity...

6. You can use the root/ratio test.
Define $x_n = n!$.
Since $\lim \frac{x_{n+1}}{x_n} = +\infty$ it means $\lim |x_n|^{1/n} = \lim (n!)^{1/n} = \infty$.

7. Yes,I know that ratio test helps better;but I was having trouble with the question I posted.In the book the problem was solved using Cauchy's root test.