$$\sum {z^n\over n!} $$
I used Alembert's Ration test and get $$\lim_{n \infty}{u_n\over u_{n+1}}={n+1\over z}$$ As this tends to $\infty>1$ can i say that the given series is convergent for all values of $z$ ?
Note : z is a complex number
Asked
Active
Viewed 194 times
3 Answers
1
Yes. A good thing to remind is the proof of this property, which in your case writes:
$$ \frac{u_{n+1}}{u_n} = \frac z{n+1} \implies \left|\frac{u_{n+1}}{u_n}\right| \le 1/2 <1 $$when $n$ is big enough, and then $$ |u_n| \le C2^{-n} $$for every $n$, for a certain $C>0$. Hence the series is (absolutely) convergent.
mookid
- 28,236
-
The problem is actually a part of a problem from complex analysis. It says that the given series is convergent for $|z|<\infty$ , is it just to avoid $\infty / \infty$ – Aman Mittal Jun 06 '14 at 08:09
0
You're wrong. The D'Alemebert Ratio test considers $$ \lim_{n\rightarrow+\infty}\frac{u_{n+1}}{u_n} $$ NOT $$ \lim_{n\rightarrow+\infty}\frac{u_{n}}{u_{n+1}} $$
so correct limit to consider is $$ \frac{u_{n+1}}{u_n}=\frac{z}{n+1}\stackrel{n\rightarrow+\infty}{\longrightarrow}0\;\;\;\forall z\in\mathbb C $$
Hence your power series converges on the whole complex plane.
To be precise: the series converges to a well known entire function: actually we have that $$ \sum_{n=0}^{+\infty}\frac{z^n}{n!}=e^z\;\;\;\forall z\in\mathbb C. $$
Joe
- 11,745
-1
The summation of the given series can be given as $e^z$ and hence the given series is convergent.
user3628041
- 307
-
-
The first part is true, the second part is restating the question. – Bennett Gardiner Jun 06 '14 at 08:05
-
1I do not understand the "recommend deletion" votes here. It is an attempt to answer the question, albeit not really correct. Surely downvoting is the correct response here, not voting to delete! – user1729 Jun 06 '14 at 08:57