I firstly found the simplified form of $\frac{a_{n+1}}{a_n} = |x|\cdot|x^n|$ and used this to establish the end points $-1\lt x\lt 1$.

I then tested the end points by finding the limit to infinity of $(-1)^{n!}$ and $1^{n!}$. By doing this I established that (by the divergence test) at neither end-point the series converged.

Therefore I concluded that the interval of convergence is $(-1,1)$.

I'm not sure if this is correct. If it's wrong can you please tell me where I went wrong (but don't tell me the full answer please!).

  • 1
    $\begingroup$ $\displaystyle\frac{x^{(n+1)!}}{x^{n!}}=|x|\cdot |x^n|$? Nope! The rule is $\displaystyle\frac{x^a}{x^b}=x^{a-b}$... $\endgroup$ – anon May 26 '15 at 21:11
  • $\begingroup$ Try comparing the behavior of $\sum_{n = 0}^\infty x^{n!}$ with $\sum_{n = 0}^\infty x^n$. $\endgroup$ – Austin Mohr May 26 '15 at 21:15

Your use of the ratio test is fine--clearly, it won't converge outside of $[-1,1]$, and you are correct in asserting that it doesn't converge for $x=-1$ and $x=1$.

You probably still have to prove that it converges for $x\in (-1,1)$, but you can use the comparison test against the series $a_n = x^n$ for that.

  • $\begingroup$ ... Except that the ratio is actually $x^{nn!}$ and not $|x||x^n|$. $\endgroup$ – AlexR May 26 '15 at 21:25
  • $\begingroup$ Ah, right. But otherwise correct. $\endgroup$ – Giuseppe May 27 '15 at 22:49

Split this up into three parts:

1). $x\in (-1,1)$

2). $x\notin (-1,1)$

3). $x=1$ or $x=-1$.

1). In this case your general term is $\leq $ $x^n$. On comparison with the geometric series, your series then converges.

2) and 3). In these cases, the series diverges, because the general term does not go to zero as $n\rightarrow \infty$.


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