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Let consider the power series $$ \sum_{k=1}^\infty a_kx^k. $$ We know that it converges when $$\lim_{n\to \infty }\sqrt[n]{|a_n|}|x|< 1\quad\text{or}\quad |x|< \frac{1}{\lim_{n\to \infty }\sqrt[n]{|a_n|}}$$

But D'Alembert test tells us that it converges if $$|x|< \frac{1}{\lim_{n\to \infty }\left|\frac{a_{n+1}}{a_n}\right|},$$ so since the radius of convergence is unique, we should have that $$\lim_{n\to \infty }\sqrt[n]{|a_n|}=\lim_{n\to \infty }\left|\frac{a_{n+1}}{a_n}\right|.$$

I tried to prove it, but this result looks strange to me. So if it doesn't work always, how can we have those to limit as radius of convergence? I guess the equality of those two limit should be true most of the time. So under what conditions is it true?

Martin Sleziak
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user330587
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    If both limits exist, then this is true. I think it might be true if either one of the limits exists, but I'm not sure. – Thomas Andrews Jun 14 '17 at 16:29
  • In the simplest case, if $a_n$ is a geometric sequence, they clearly agree. – Arthur Jun 14 '17 at 16:30
  • @ThomasAndrews Can you show it ? – stity Jun 14 '17 at 16:31
  • Can I show what? @stity – Thomas Andrews Jun 14 '17 at 16:33
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    See THIS ANSWER, which proves that $$\liminf \frac{a_{n+1}}{a_n}\le \liminf (a_n)^{1/n}\le \limsup (a_n)^{1/n}\le \limsup \frac{a_{n+1}}{a_n}$$Clearly, if $\lim \frac{a_{n+1}}{a_n}$ exists then so does $\lim (a_n)^{1/n}$. The converse is NOT generally true. – Mark Viola Jun 14 '17 at 16:34
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    You need to change $\le$ to $<.$ – zhw. Jun 14 '17 at 17:09
  • See this question and other posts linked there. – Martin Sleziak Jun 15 '17 at 13:26
  • Perhaps also these posts can be of interest: http://math.stackexchange.com/questions/76743/limit-of-fraca-n1a-n http://math.stackexchange.com/questions/287932/convergence-of-ratio-test-implies-convergence-of-the-root-test http://math.stackexchange.com/questions/1079637/how-to-show-that-lim-n-to-infty-a-n1-n-l http://math.stackexchange.com/questions/2045648/lim-n-to-infty-a-n-frac1n-l-implies-lim-n-to-infty-frac – Martin Sleziak Jun 15 '17 at 13:30
  • This is the famous fact: Root test is stronger than Ratio test, in particular if the limit |a_n+1/a_n| exists then limit |a_n|^1/n also exists and thus is equal to the limit superior – Anshuman Agrawal Mar 15 '23 at 16:01

2 Answers2

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If the limit $$ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|, $$ exists in $[0,\infty]$, then so does $$ \lim_{n\to\infty}\sqrt[n]{|a_n|} $$ and the limits are equal.

However, if the second limit exists, then the first one DOES NOT HAVE to exist. For example $$ a_n=\left\{\begin{array}{lll} 1 & \text{if} & n\,\, \text{odd},\\ 2 & \text{if} & n\,\, \text{even}, \end{array}\right. $$ then $$ \lim_{n\to\infty}\sqrt[n]{|a_n|}=1, $$ while the first limit does not exist.

In general, $$ \liminf_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\le \liminf_{n\to\infty}\sqrt[n]{|a_n|}\le \limsup_{n\to\infty}\sqrt[n]{|a_n|}\le \limsup_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| \tag{1} $$

Note. A consequence of $(1)$ is that, if both limits exist, then they are equal. In fact, if the ratio limit exists, then root limit also exists, and their are equal. However, it is possible for the root limit to exist, while the ratio one to not exist.

Yiorgos S. Smyrlis
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  • So how can both can be defined as radius of convergence if both con not be the same ? – user330587 Jun 15 '17 at 06:48
  • @user330587 If both limits exist in $[0,\infty]$, then they are equal, and hence the define the same radius of convergence. The problem is that it is possible for the ratio limit to not exist, while at the same time the root limit does exist. – Yiorgos S. Smyrlis Jun 15 '17 at 12:41
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If $\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|$ exists, so does $\lim_{n\rightarrow\infty}\sqrt[n]{|a_n|}$ and has the same value: because then, the geometric mean $\sqrt[n]{\prod^n_{k=0}\left|\frac{a_{k+1}}{a_k}\right|}$ has the same limit. It does not work in the other direction.