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I am learning analysis on my own and I am puzzled with the following question.

Consider the series $$\frac{1}{2}+\frac{1}{3}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^3}+\cdots$$ Indicate whether this series converges or diverges.

The following is what I was able to get.

First I noticed that $$a_n = \frac{1}{2^n}+\frac{1}{3^n}$$ would be a simple and nice form of this sequence, so according to the ratio test, $$\lim_{n \to \infty} \frac{1}{6} \left( \frac{3+2(\frac{2}{3})^n}{1+(\frac{2}{3})^n}\right) = \frac{1}{2}$$ therefore the series must converge.

However, I later understood the concept of upper an lower limits a little, so taking $$ a_n = \begin{cases} \frac{1}{2^n}, & \text{if $n$ is odd} \\ \frac{1}{3^{n-1}}, & \text{if $n$ is even} \\ \end{cases}$$ I figured out the following $$\lim_{n \to \infty} \frac{1}{2}\left(\frac{2}{3}\right)^n = 0$$ when $n$ is odd and $$\lim_{n \to \infty} \frac{1}{2}\left(\frac{3}{2}\right)^n = +\infty$$ when $n$ is even.

Through this experience I realized and really understood how a subsequence of a sequence may have multiple limits, and the supremum and the infimum of those limits give us the upper and lower limits, respectively. But the book does something different than I did, and I would like to have the following questions answered.

1),$\quad$ From my understanding, the ratio test tells us that $\Sigma a_n$ diverges when the upper limit is greater than 1. Clearly one of the limits goes to $+\infty$, so why cannot we conclude that the series diverge ?

2),$\quad$ I learned that by choice of the subsequence, the limit may or mayn't be different. In the case that it is different, how do we know what is the supremum and infimum? It would be clear that if they were $\pm \infty$ but in my case, why is it guaranteed that there is no other subsequence that has a negative limit ? Is there an algebraic way to guarantee them ? Or do we always have to conceptually understand all of the subsequential limits.

3),$\quad$ The book says the ratio test is inconclusive, which turns me back to question 1), but it used the ratio test to determine that the series converges. According to my even-odd wise definition of $a_n$, the "larger limit" (I am purposefully not saying supremum because I don't know how it's guaranteed that it is the supremum) using the ratio test can be found as $$\lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \sqrt[n]{\frac{1}{2^n}} = {1 \over 2}$$ But the book says $$\lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \sqrt[2n]{\frac{1}{2^n}} = {1 \over \sqrt{2}}.$$ Where did the $2n$ come from ?

Can someone help me out ?

Edit:Thanks for helping me out everyone! I really appreciate it.

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  • $\begingroup$ I do not understand why do you have the expression "$\lim_{n \to \infty} \frac{1}{2}(\frac{3}{2})^n = +\infty$ when $n$ is even." This is not related to the original problem that you have posed. The subsequence $\frac{1}{3^{n-1}}$ converges to $0$ as well? $\endgroup$ – Lord Soth May 25 '13 at 3:22
  • $\begingroup$ Hint: ratio test $\endgroup$ – hyg17 May 25 '13 at 3:24
  • $\begingroup$ Oh ok, you are doing the ratio test again. $\endgroup$ – Lord Soth May 25 '13 at 3:26
  • $\begingroup$ It reminds me the series: $1+1-1+1-1+\cdots$. You can consider it as $1+(1-1)+(1-1)+\cdots=1$ or $(1+1)-(1-1)-(1-1)-\cdots=2$. So, as it's unclear specified and can be converged to different values, the series actually diverges. $\endgroup$ – Stan May 25 '13 at 3:35
  • $\begingroup$ Are you sure ? This series should converge. $\endgroup$ – hyg17 May 25 '13 at 3:37
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By pairing adjacent terms, you’ve actually changed the series with which you’re working: you’re working with $\sum_{n\ge 1}\left(\frac1{2^n}+\frac1{3^n}\right)=\sum_{n\ge 1}\frac{2^n+3^n}{6^n}$, the series whose $n$-th term is $\frac{2^n+3^n}{6^n}$. The ratio test works fine on this series:

$$\begin{align*} \lim_{n\to\infty}\frac{\frac{2^{n+1}+3^{n+1}}{6^{n+1}}}{\frac{2^n+3^n}{6^n}}&=\lim_{n\to\infty}\frac{2^{n+1}+3^{n+1}}{6(2^n+3^n)}\\ &=\lim_{n\to\infty}\frac{2^{n+1}}{6(2^n+3^n)}+\lim_{n\to\infty}\frac{3^{n+1}}{6(2^n+3^n)}\\ &=\lim_{n\to\infty}\frac{2^n}{3(2^n+3^n)}+\lim_{n\to\infty}\frac{3^n}{2(2^n+3^n)}\\ &=\lim_{n\to\infty}\frac1{3\left(1+\left(\frac32\right)^n\right)}+\lim_{n\to\infty}\frac1{2\left(\left(\frac23\right)^n+1\right)}\\ &=0+\frac12\\ &=\frac12\;, \end{align*}$$

which is essentially the calculation that you made. However, if you apply the ratio test to the original series, you find that if the $n$-th term is $a_n$, then

$$a_n=\begin{cases} \frac1{2^{(n+1)/2}},&\text{if }n\text{ is odd}\\\\ \frac1{3^{n/2}},&\text{if }n\text{ is even}\;, \end{cases}$$

so that

$$\begin{align*} \frac{a_{n+1}}{a_n}&=\begin{cases} \frac{1/3^{(n+1)/2}}{1/2^{(n+1)/2}},&\text{if }n\text{ is odd}\\\\ \frac{1/2^{(n+2)/2}}{1/3^{n/2}},&\text{if }n\text{ is even} \end{cases}\\\\ &=\begin{cases} \left(\frac23\right)^{(n+1)/2},&\text{if }n\text{ is odd}\\\\ \frac12\left(\frac32\right)^{n/2},&\text{if }n\text{ is even}\;. \end{cases} \end{align*}$$

Thus, $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$ does not exist: the terms with odd indices are approaching $0$, but those with even indices are increasing without bound. Since the limit does not exist, the ratio test is inconclusive.

The ratio test applied to your modified series gives the correct answer because the original series is absolutely convergent; this permits you to combine the adjacent terms as you did, but you since you don’t know ahead of time that the series converges, you can’t make use of the fact to show that it converges.

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  • $\begingroup$ Finally someone that makes sense. Thanks. But that rises another question in me. How do I know if I "changed" the series ? How is it different from having, say, $1/4 + 1/4 + 1/6 + 1/6 + 1/8 + 1/8 + 1/18 + 1/18 + ...$ ? $\endgroup$ – hyg17 May 25 '13 at 3:41
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    $\begingroup$ @hyg17: Any time you rearrange the order of the terms or combine some of them, you’re technically changing the series. A finite number of such changes is permissible, since convergence depends only on what happens ‘in the long run’, but you made infinitely many changes by combining infinitely many pairs of terms. In fact one can prove that if the terms are all positive, the kind of combining that you did is harmless, but this is something that has to be proved. $\endgroup$ – Brian M. Scott May 25 '13 at 3:45
  • $\begingroup$ :That is great to know. I still haven't seen such proof or as a matter of fact, never even thought about that so it threw me off. Thank you . $\endgroup$ – hyg17 May 25 '13 at 4:55
  • $\begingroup$ @hyg17: You’re welcome! $\endgroup$ – Brian M. Scott May 25 '13 at 4:56
  • $\begingroup$ @Brian: "you can’t make use of the fact to show that it converges" is not true. You should make use of that fact, and indeed you should do it properly. The series in question has positive terms, so that fact proves that the series is absolutely convergent. $\endgroup$ – Hendrik Vogt May 25 '13 at 10:32
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The even terms are a geometric series, with sum $\frac{1/3}{1-1/3}=\frac{1}{2}$. The odd terms are a geometric series, with sum 1. Because all the terms are positive, the series is absolutely convergent, so you may rearrange the odd terms to go first, then the even terms. Hence the overall series is convergent, with sum $\frac{3}{2}$.

In answer to your specific questions, the ratio test is indeed inconclusive in this case, because the limit of the ratios does not exist. That doesn't mean the series diverges, it means the test doesn't help. The supremum is infinite, the infimum is finite, so even with that refinement of the ratio test it's inconclusive.

With regards to your third question, the n-th root test is about specifically the n-th term. In this series, $\frac{1}{2^n}$ is the 2n-th term. If you group the terms pairwise as you did, then you need to take the limit of $\sqrt[n]{\frac{1}{2^n}+\frac{1}{3^n}}$.

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  • $\begingroup$ I appreciate your help, but I know how to find the sum and that's not what I'm asking. $\endgroup$ – hyg17 May 25 '13 at 3:16
  • $\begingroup$ Could you explain me why the limits of the ratios does not exist ? Also, I tried to find the limit of $\sqrt[n]{\frac{1}{2^n}+\frac{1}{3^n}}$ but I couldn't find it... $\endgroup$ – hyg17 May 25 '13 at 3:26
  • $\begingroup$ The limit of the ratios does not exist for the same reason that the following sequence does not have a limit: $1/1, 1, 1/2, 2, 1/3, 3, 1/4, 4, \ldots$. $\endgroup$ – vadim123 May 25 '13 at 3:32
  • $\begingroup$ Uhm, so the infimum of the ratio and the supremum of the ratio doesn't match. Is that what's going on ? $\endgroup$ – hyg17 May 25 '13 at 3:36
  • $\begingroup$ @hyg17, that's correct, so the limit does not exist. $\endgroup$ – vadim123 May 25 '13 at 3:38
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The ratio test says that IF a certain limit is less than $1$ then a certain series converges and IF that limit is more than $1$ then that series diverges. The ratio test does not say whether the series converges or diverges in cases in which that limit does not exist (nor when it's equal to $1$). In this case, the limit in the ratio test does not exist, so the ratio test (in that form, at least) doesn't tell you whether the series converges or not. Thus the ratio test does not give you any conclusion that can contradict any other conclusion about whether the series converges.

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  • $\begingroup$ Could you explain me how the limit doesn't exist ? I thought it was 0. $\endgroup$ – hyg17 May 25 '13 at 3:32
  • $\begingroup$ You're looking at $\dfrac{a_{n+1}}{a_n}$. For one value of $n$ this is $\dfrac{3^{n-1}}{2^{n+1}}\to\infty$ and in the next it's $\dfrac{2^n}{3^n}\to0$. If one subsequence approaches $\infty$ and another approaces $0$, the the limit of the sequence as a whole does not exist. $\endgroup$ – Michael Hardy May 25 '13 at 3:42
  • $\begingroup$ Thanks ! That explains a lot. $\endgroup$ – hyg17 May 25 '13 at 4:57
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1) ... is divergent when the lower limit of $a_{n+1}$ to $a_n$ is greater than $1$. If it, indeed, says "the upper" there, it is a misprint.

2) I cannot understand what you are asking. Can you restate the question in a clearer way?

3) Formally $1/2^n$ stands in the $2n-1$-th position in the original series, so it should really be $\lim_{n\to\infty}\sqrt[2n-1]{1/2^n}=\frac{1}{\sqrt{2}}$. Of course, if you remove half of the terms, you change the indexation and, as a consequence, the quantity in the root test.

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  • $\begingroup$ 2), What I'm saying is, I got 3 different limits. 1/2, 0 and $\infty$. How do I know which one ? $\endgroup$ – hyg17 May 25 '13 at 3:42
  • $\begingroup$ 3), That makes so much sense. Thanks ! $\endgroup$ – hyg17 May 25 '13 at 3:43

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