find the radius of convergence of the power series $\sum_{n=0}^{\infty} z^{n!}$.

Question

My attempt : $\sum_{n=0}^\infty z^{n!}$ is a power series of the form $\sum_{n=0}^\infty {c_n z^n}$ with $$ c_n=\begin{cases} 1 & \text{if $n=k!$ for some integer $k$,}\\ 0 & \text{else}. \end{cases} $$

Any hints/idea/solution will be appreciated

thanks u

Also: https://math.stackexchange.com/q/491088/42969, https://math.stackexchange.com/q/2130255/42969, https://math.stackexchange.com/q/595501/42969 – all found with Approach0 — Martin R, Feb 02 '19 at 07:52

score 4 · Answer 1 · answered Feb 02 '19 at 08:10

The radius has to be 1. If $\vert z \vert \ge 1$, then the series can't converge because its terms don't approach $0$. If $\vert z \vert \lt 1$, then the series is smaller in absolute value than $\sum_{n=0}^\infty {z^n}$, which converges to $1/(1-z)$ when $\vert z \vert \lt 1$.

score 3 · Accepted Answer · answered Feb 02 '19 at 06:35

3

The series is $\sum_k a_kz^k$ where $a_1=2$, $a_k=1$ if $k=n!$ for some $n\ge2$, and $a_k=0$ otherwise.

answered Feb 02 '19 at 06:35

Hagen von Eitzen

374,180

if $k =1 $ how $ a_1 = 2 ?$ can u elaborate more – jasmine Feb 02 '19 at 06:39
1

@jasmine : because $0!=1!=1$ – Ayoub Feb 02 '19 at 06:41
@Ayoub thanks i missed that logics – jasmine Feb 02 '19 at 06:43

score 1 · Answer 3 · 2019-02-02T11:30:06.893

1

Use the Cauchy-Hadamard theorem ($r=\frac1{\limsup_{n\to\infty} \mid a_n\mid^{\frac1n}}$) and the fact that $\begin{cases}a_k=1, k=n!\\a_k=0,\text{otherwise}\end{cases}$. You'll get $r=1$.

Or Abel's test, to get $r\ge1$. Then $r\le1$ since $a_n\not\to0$.

edited Feb 02 '19 at 11:30

answered Feb 02 '19 at 11:12

find the radius of convergence of the power series $\sum_{n=0}^{\infty} z^{n!}$.

3 Answers3