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

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##.

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!

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.

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?

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##.

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.

I analyze whole question and I think here limit will not exist.