0

Let me explain. So take the series

  • $\sum_{n=0}^\infty (-2)^nn^4(x-1)^n$

In order to find the radius of convergence for this power series, I used the ratio test

  • $\lim_{n\to \infty} |\frac{2(n+1)^4(x-1)^{n+1}}{2n^4(x-1)^n}|$

and got $|x| < 2$ but then the radius of convergence was $R = \frac{1}{2}$

Is the radius of convergence $R$ obtained by putting it over 1? I'm really confused

Edit: Calculated limit wrong. Sorry

Bill Cogn
  • 81

1 Answers1

2

When you calculate the limit, you get $$L = \lim_{n\to\infty} \left|\frac{2(n+1)^4(x-1)^{n+1}}{2n^4(x-1)^n}\right| = 2|x-1|.$$ The ratio test gives convergence so long as $L<1,$ so you need $2|x-1|<1,$ meaning $|x-1|<\frac{1}{2}.$ This exactly says that the radius of convergence is $\frac{1}{2}.$

D. Brogan
  • 3,427
  • 10
  • 20
  • 2
    @BillCogn: I don't think that was where you went wrong! You seem to think that the radius of convergence is $\lim\left |\frac{a_{n+1}}{a_n}\right |$ if this limit exists; whereas in fact it is $\lim\left |\frac{a_n}{a_{n+1}}\right |$. – TonyK Jan 05 '24 at 15:59
  • @TonyK Ahh that's it – Bill Cogn Jan 05 '24 at 16:03
  • @TonyK Is there a reason why it happens to be the reciprocal of the ratio test? – Bill Cogn Jan 05 '24 at 16:08
  • 1
    To understand "the reason", try the ratio test on a geometric series. (As in the hint by Amateur-Algebraist.) – GEdgar Jan 05 '24 at 16:15
  • @GEdgar: That doesn't help, if by "geometric series" you mean $\sum_{n=1}^\infty x^n$ $-$ the limits of $\left |\frac{a_{n+1}}{a_n}\right |$ and $\left |\frac{a_n}{a_{n+1}}\right |$ are the same. – TonyK Jan 05 '24 at 16:17
  • Try the geometric series $\sum r^n x^n$, where $r>0$ is a given constant. – GEdgar Jan 05 '24 at 16:18
  • That's better :-) – TonyK Jan 05 '24 at 16:18