# Rudin Series ratio and root test.

In Rudins Principles of Mathematical Analysis he says consider the following series

$$\frac 12 + \frac 13 + \frac 1{2^2} + \frac 1{3^2} + \frac 1{2^3} + \frac 1{3^3} + \frac 1{2^4} + \frac 1{3^4} + \cdots$$

for which

$$\liminf \limits_{n \to \infty} \dfrac{a_{n+1}}{a_n} = \lim \limits_{n \to \infty} \left( \dfrac {2}{3} \right)^n =0,$$

$$\liminf \limits_{n \to \infty} \sqrt[n]{a_n} = \lim \limits_{n \to \infty} \sqrt[2n]{\dfrac{1}{3^n}} = \dfrac{1}{\sqrt{3}},$$

$$\limsup \limits_{n \to \infty} \sqrt[n]{a_n} = \lim \limits_{n \to \infty} \sqrt[2n]{\dfrac{1}{2^n}} = \dfrac{1}{\sqrt{2}},$$

$$\limsup \limits_{n \to \infty} \dfrac{a_{n+1}}{a_n} = \lim \limits_{n \to \infty} \dfrac 12\left( \dfrac {3}{2} \right)^n =+\infty,$$

The root test indicates convergence; the ratio test does not apply.

In the book he defines the root and ratios test for the lim sup. I am not exactly sure how he goes from the lim sup to the lim and also why there is a $2n$ (which I assume comes from even terms of the sequence) in the root test. Also why is he checking the lim inf? I believe that my understanding of lim sups and infs are not well developed or I would probably understand what’s going on.

Also how does he get the terms that he is taking the limit of. A nudge in the right direction to figure this out would be much appreciated. Thank you!!

• One of the expressions is not quite right, which one depends on whether $n$ starts at $0$ or at $1$. Doesn't matter, the $2n+1$-th root and the $2n$-th root of $\frac{1}{2^n}$ have the same limit. – André Nicolas Oct 25 '12 at 2:17
• The only thing that was wrong was the 2/3 in the last limit. I changed it to the 3/2 which it should be. – Differintegral Oct 25 '12 at 2:21
• I was commenting about the Root Test expression. If we want to take the $k$-th root of the $k$-th term, in this case we will alternately be looking at the $2n$-th root and the $2n+1$-th root of an $n$-th power. – André Nicolas Oct 25 '12 at 2:26
• I also have some confusion regarding this concept. Would like to clarify about the same. Rudin in his introduction to ratio test says that a series converges if the limsup is less than one. It diverges if the ratio is more than 1 for all n greater than a N. I guess the crux of the matter is that limsup is not used in both the cases. If it were it would cover all the cases. Wikipedia says that series converges if limsup less than one and it diverges if liminf more than 1. Thus , the two conditions dont cover all the cases. And the ratio test is not helpful in cases where limsup >1 and liminf<1. – ameyask Jan 29 '14 at 12:53
• contd. Which is what is happening in the example of this question. Is my understanding correct ? I would have posted a new question , but this was already asked. And my query is just about a confirmation hence asking in comment. I hope its ok. – ameyask Jan 29 '14 at 12:54

For definiteness, call the terms of our sequence $a_1,a_2,a_3,\dots$. A similar analysis with minor differences of detail can be made if we call the first term of our sequence $a_0$.

Note that for $n=1,2,3,\dots$ we have $a_{2n-1}=\dfrac{1}{2^n}$ and $a_{2n}=\dfrac{1}{3^n}$.

The $k$-th root of the $k$-th term is "small" when the $k$-th term is a power of $\dfrac{1}{3}$. The $k$-th root of the $k$-th term is "large" when the $k$-th term is a power of $\dfrac{1}{3}$.

More precisely, $\liminf \sqrt[k]{a_k}=\lim\inf \sqrt[2n]{\frac{1}{3^n}}=\dfrac{1}{3}$. For even $k$ the $k$-th root is constant.

Also, $\limsup\sqrt[k]{a_k}=\liminf\sqrt[2n-1]{\dfrac{1}{2^n}}$. But $$\sqrt[2n-1]{\dfrac{1}{2^n}}=\left(\frac{1}{2^n}\right)^{1/(2n-1)}=\left(\frac{1}{2^n}\right)^{2n/(2n(2n-1))}=\left(\frac{1}{\sqrt{2}}\right)^{2n/(2n-1)}.$$ The expression on the right has limit $\dfrac{1}{\sqrt{2}}$.

That takes care of one of the gaps.

For the Ratio Test, we are interested in the behaviour of $\left|\dfrac{a_{k+1}}{a_k}\right|$.

Let $k$ be odd, say $k=2n-1$. Then $a_k=\dfrac{1}{2^n}$. And $a_{k+1}=a_{2n}=\dfrac{1}{3^n}$. It follows that $$\frac{a_{k+1}}{a_k}=\frac{a_{2n}}{a_{2n-1}}=\left(\frac{2}{3}\right)^n.$$ This has very pleasant behaviour for large $n$, indeed for any $n$: it is safely under $1$, indeed has limit $0$.

Now let $k$ be even, say $k=2n$. Then $a_k=\dfrac{1}{2^n}$. and $k+1=2n+1$. The $2n+1$-th term of our sequence is $\dfrac{1}{2^{n+1}}$. It follows that in the case $k=2n$ we have $$\frac{a_{k+1}}{a_k}=\frac{a_{2n+1}}{a_{2n}}=\frac{\frac{1}{2^{n+1}}}{\frac{1}{3^n}}=\frac{1}{2}\left(\frac{3}{2}\right)^n.$$

This unfortunately behaves badly for large $n$: we would like it to be safely under $1$, and it is very much over.

The limit of the ratios $\dfrac{a_{k+1}}{a_k}$ does not exist. The ratios do not (uniformly) blow up, since for $k$ odd, the ratios approach $0$. The ratio behaves very nicely at odd $k$, and very badly at even $k$. So the Ratio Test is inconclusive. The bad behaviour prevents us from concluding convergence. But the good behaviour prevents us from concluding divergence.

• Thank you for all the work you did to show me. It is making more sense now but I am still curious about something. Both the root test and the ratio test only need the lim sup and the possibly the ratio $|a_{n+1}/a_n|$ on it's own. Why doesn he look at the lim inf in this case? I feel like there is one major concept that I am missing here. Also is the reason we can turn the lim sup and lim inf into just the limit is because the limit exists and therefore we have that lim inf = lim sup = lim? Of course in the case of $+\infty$ it exists in the extended reals. – Differintegral Oct 25 '12 at 4:48
• Second question: yes, when we evens and odds, the limsups are actual limits. In the Ratio Test, we are interested in limsup because, at the heart of things, the Ratio Test is a comparison with a geometric series. If, in the long run, the ratio (of absolute values) is $\lt r\lt 1$, the comparison with a geometric series with common ratio $r$ yields convergence. (to be continued) – André Nicolas Oct 25 '12 at 5:11
• (Cont.) The Root Test is also fundamentally a comparison with a geometric series. If $\sqrt[n]{a_n}$ is ultimately $\lt r\lt 1$, then ultimately $a_n\lt r^n$. The Root Test doesn't care if the $a_n$ bounce around a bit, as long as they go down fast enough. The Ratio Test kind of insists that (after a while) we steadily go down fastish, term after term. – André Nicolas Oct 25 '12 at 5:12
• so we conclude that the series converges, is that right? – emmy Apr 26 '17 at 2:15