How does the ratio test fail and the root test succeed here?

  • #1
psie
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I'm studying Spivak's Calculus, chapter 23, problem 7. There he introduces the root test and he gives an example of a series for which the ratio test fails but the root test works. I struggle with verifying this.
The series that is given is $$\frac12+\frac13+\left(\frac12\right)^2+\left(\frac13\right)^2+\left(\frac12\right)^3+\left(\frac13\right)^3+\ldots.$$ Now, it's easy to see these are two separate geometric series, however, Spivak claims the ratio test fails because the ratio of successive terms does not approach a limit. I have figured out that $$a_{2n-1}=\left(\frac12\right)^{n},\quad a_{2n}=\left(\frac13\right)^{n},\quad n=1,2,\ldots.$$ For the ratio test, we should have that the fraction ##\left|\frac{a_{n}}{a_{n-1}} \right| ## approaches some finite limit or diverges to infinity. In this case, if ##n## is even we have that the fraction is ##\left(\frac23\right)^n\to 0## as ##n\to\infty##. If ##n## is odd, we have ##\left(\frac32\right)^n\frac13\to\infty## as ##n\to\infty##. This behavior confuses me. What can we conclude from this?

The root test given in the exercise is that if ##a_n\geq0## and ##\lim\limits_{n\to\infty}\sqrt[n]{a_n}=r##, then ##\sum_{n=1}^\infty a_n## converges if ##r<1## and diverges if ##r>1##. Apparently this test should work, but I do not see how when the even indexed subsequence and the odd indexed subsequence seem to converge to different limits, namely ##\frac13## and ##\frac12##.
 
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  • #2
How does Spivak define the root test? The following definition uses the ##\lim \sup##.

https://en.wikipedia.org/wiki/Root_test

That said, it seems more logical to me to use the convergence of both subseries. It can't be hard to prove that if both ##\sum_{n = 1}^{\infty} a_n## and ##\sum_{n = 1}^{\infty} b_n## converge, then ##a_1 + b_1 + a_2 + b_2 + a_3 + b_3 \dots## must also converge.

PS and in this case, it's not hard actually to compute the limit of the series!
 
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  • #3
The root test is stronger than the ratio test, at least if ##\limsup##-version is used. It is based on this inequality: ##\liminf(a_{n+1}/a_n) \le \liminf a_n^{1/n} \le \limsup a_n^{1/n} \le \limsup(a_{n+1}/a_n)##, if every ## a_n## is a positive real number.
 
  • #4
PeroK said:
How does Spivak define the root test?
Spivak defines the root test first as I defined it above, without the ##\limsup##. Then, however, he adds that we can replace the limit with ##\limsup## and calls it the "delicate root test". He probably means that if we use the root test with ##\limsup## on the series above, then we get that ##\limsup\limits_{n\to\infty}|a_n|^{1/n}=\frac12##, which is less than ##1##, so we have convergence.

PeroK said:
That said, it seems more logical to me to use the convergence of both subseries. It can't be hard to prove that if both ##\sum_{n = 1}^{\infty} a_n## and ##\sum_{n = 1}^{\infty} b_n## converge, then ##a_1 + b_1 + a_2 + b_2 + a_3 + b_3 \dots## must also converge.
I agree it would be easier to use the convergence of both subseries, but I'm trying to understand what Spivak means by the ratio test failing. We have that ##\left|\frac{a_{n}}{a_{n-1}} \right|## converges to ##0## if ##n## is even and ##\infty## if ##n## is odd, as ##n\to\infty##. Can we conclude from this that ##\lim\limits_{n\to\infty}\left|\frac{a_{n}}{a_{n-1}} \right|## does not equal a finite number or infinity?
 
  • #5
psie said:
Spivak defines the root test first as I defined it above, without the ##\limsup##. Then, however, he adds that we can replace the limit with ##\limsup## and calls it the "delicate root test". He probably means that if we use the root test with ##\limsup## on the series above, then we get that ##\limsup\limits_{n\to\infty}|a_n|^{1/n}=\frac12##, which is less than ##1##, so we have convergence.I agree it would be easier to use the convergence of both subseries, but I'm trying to understand what Spivak means by the ratio test failing. We have that ##\left|\frac{a_{n}}{a_{n-1}} \right|## converges to ##0## if ##n## is even and ##\infty## if ##n## is odd, as ##n\to\infty##. Can we conclude from this that ##\lim\limits_{n\to\infty}\left|\frac{a_{n}}{a_{n-1}} \right|## does not equal a finite number or infinity?
The ratio test is inconclusive, as every other term in the sequence of ratios is ##\frac 1 2(\frac 3 2)^n##.
 
  • #6
psie said:
I agree it would be easier to use the convergence of both subseries, but I'm trying to understand what Spivak means by the ratio test failing. We have that ##\left|\frac{a_{n}}{a_{n-1}} \right|## converges to ##0## if ##n## is even and ##\infty## if ##n## is odd, as ##n\to\infty##. Can we conclude from this that ##\lim\limits_{n\to\infty}\left|\frac{a_{n}}{a_{n-1}} \right|## does not equal a finite number or infinity?

The correct conclusion is that this limit does not exist.
 
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  • #7
I analyze whole question and I think here limit will not exist.
 

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