I just wondering radius of convergence following series $$ \sum_{n=1}^{\infty}n! x^{n!} \\ $$ My 1st attempt is 'root test' $$ \sqrt[n!]{|a_{n!} |} =\sqrt[n!]{|n! |} =\sqrt[t]{t} \rightarrow 1 $$ So, I thoungt radius of convergence is '1'

Is this right answer?

$$ $$

$$ $$

And 2nd question!!

My friend solve this problem wiht "ratio test"

He said

$$ |\frac{(n+1)!x^{(n+1)!}}{n!x^{n!}}| \le 1 \implies |(n+1)x^{n \cdot n!}|\le1 \implies |x|\le|\frac{1}{n+1}|^{\frac{1}{n\cdot n!}} \rightarrow1 $$

So, radius of convergence is '1'

Is this right answer?

$$ $$ I've solved problem of this type through $$ |\frac{a_{n+1}x^{n+1}}{a_n x^n}|\le1 $$ So, I am uncertain for my friend's solution.

BECAUSE for the series, $$ a_2=0, a_3=0, a_k=0 ( k \not=n!) $$

thus there will be $$ \frac{0}{0} $$

$$ $$

Please give me some advice.

Thanks in advance.

  • $\begingroup$ Yes, $R = 1$. Cauchy-Hadamard formula would also work. Just do the calculations thinking of $\sum_{n=1}^\infty nx^n$ $\endgroup$ – Paul Hurst Jun 8 '14 at 7:19
  • $\begingroup$ I modified 2nd question. Does C-H formula work well for case 0/0? $\endgroup$ – user143993 Jun 8 '14 at 7:30
  • $\begingroup$ I don't think the root test that you demonstrated is correct. Also are you sure that the power of $x$ is $n!$? $\endgroup$ – jdoicj Jun 8 '14 at 7:39
  • $\begingroup$ "the power of x is n!".... umm.. SUM x^f(n) means 'the power of x is f(n)' isn't it? $\endgroup$ – user143993 Jun 8 '14 at 7:48
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    $\begingroup$ Your friend's solution works. The key is to NOT apply the formula for the radius of convergence. Rather, you apply the ratio or root test as if the series were NOT a power series, and then set $|L| < 1$ and solve for $x$. $\endgroup$ – Paul Hurst Jun 8 '14 at 8:15

The radius of convergence is $1$, since the sum converges for $x\in(-1,1)$, sometimes both approaches (ratio and root) will work, so it is fine that there are two methods.

Your uncertainty with $\frac{0}{0}$ is okay, because if we we had some series $\sum_{n=1}^\infty x_n$, and we knew for certain terms, where $x_n=0$, say for a set $S\subset\Bbb N$, then we would just rewrite the series as:

$\sum_{n=1}^\infty x_n=\sum_{n\in \Bbb N\setminus S}x_n$, and we would assess the remaining sequence.

Example $x_n=1/n^2$ for $n$ odd, and $x_n=0$ for $n$ even, we would just have:

$\sum_{n=1}^\infty x_n=\sum_{n\in \Bbb N\setminus 2\Bbb N}^\infty x_n$

  • $\begingroup$ Thanks for answer. You mean "reduce" the zero terms and regard as 'new' series, right? $\endgroup$ – user143993 Jun 8 '14 at 7:46
  • $\begingroup$ Indeed, if you know beforehand that some terms are zero, then you wouldn't include them. $\endgroup$ – Ellya Jun 8 '14 at 7:47
  • $\begingroup$ OK, I understand! THANK you very much. By the way, 1st solution (I did 'n! = t') is also right? $\endgroup$ – user143993 Jun 8 '14 at 7:49
  • $\begingroup$ this is fine since both values change at the same rate, but I would keep it as $n!$ personally. $\endgroup$ – Ellya Jun 8 '14 at 7:56
  • $\begingroup$ Sorry to bother you.. If keep it as n!, how can I calculate "lim{n!}^{1/n!}"? Could you give me some hint for calculating this limit? $\endgroup$ – user143993 Jun 8 '14 at 8:01

The power series here is $$ \sum_{k=0}^\infty a_kx^k $$ where $$ a_k=\left\{\begin{array}{} k&\text{if $k=n!$ for some $n\in\mathbb{Z}$}\\ 0&\text{otherwise} \end{array}\right. $$ where the non-zero terms are sparse. Several tests will fail because they reference consecutive terms (ratio test, alternating series, etc). The Root Test talks about $$ \limsup_{n\to\infty}|a_n|^{1/n} $$ This test does not care about the behavior of consecutive terms, so it is well suited to this type of series. The formula to determine the radius of convergence $$ r=\frac1{\limsup\limits_{n\to\infty}|a_n|^{1/n}} $$ is based on the Root Test.

The Comparison Test also works to show that the series converges for $|x|\le1$ since for this series $$ |a_k|\le k $$ and the series $$ \sum_{k=0}^\infty kx^k=\frac{x}{(1-x)^2} $$ converges for $|x|\lt1$.

The Term Test can also be used to show that the series does not converge for $|x|\ge1$.


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