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I have the sequence $\{ \frac{1}{1}, \frac{1}{3}, \frac{1}{2}, \frac{1}{4}, \frac{1}{3}, \frac{1}{5}, \frac{1}{4}, \frac{1}{6},...\}$. I'm having trouble coming up with an explicit description of the sequence itself, but I noticed that the sequence consists of two subsequences: $\{\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4},...\}$ and $\{\frac{1}{3},\frac{1}{4}, \frac{1}{5}, \frac{1}{6},...\}$, where the first subsequence is all of the even-indexed terms, and the second subsequence is all of the odd-indexed terms.

Since both of those subsequences converge to 0, can I say that the original sequence must also converge to 0?

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    $\begingroup$ You need to know how to generate each of the terms of the sequence(s), but yes, you can say then if both subsequences converge to the same value, then the combined sequence does the same. $\endgroup$
    – Henry
    Apr 14 at 0:07
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    $\begingroup$ Your question title says "both converge". Did you mean to say "both converge to 0"? or "converge to the same limit"? If so, yes, as the answers show. However, if they converge to different limits and each contains infinitely many of the terms, no. $\endgroup$
    – Rosie F
    Apr 14 at 8:17
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    $\begingroup$ Note: for the original sequence to converge you must include all the terms in one of the two subsequences or in both. Note: the related question for series, which are sums of sequences, gets more complicated and requires absolute convergence. $\endgroup$
    – nickalh
    Apr 14 at 10:45

3 Answers 3

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Yes, because you have accounted for all the terms in one subsequence or the other and the two limits are the same. You can go back to the $\epsilon N$ definition of a limit. If somebody gives you an $\epsilon$ and asks for an $N$ such that all the terms after the $N^{th}$ are less than $\epsilon$, you can find $N_1$ for the first subsequence and $N_2$ for the second. As in your case the corresponding terms are larger in the first subsequence you can set $N=2N_1-1$ for your reply.

The fact that the subsequences contained alternate terms from the original sequence made it easier but was not needed.

For this particular sequence, you could note that the even numbered terms are less than the preceding term and say the sequence is bounded above by $\frac 11, \frac 11, \frac 13, \frac 13, \frac 15, \frac 15 \ldots$ and that sequence converges to $0$.

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    $\begingroup$ The class is Calculus 2 so there is nothing of that technical rigor. Are there any other tools we could use to prove convergence? The text has the Monotonic Convergence Theorem but the sequence is not decreasing for all n. And since I can't explicitly define it in a single function I'm not sure if the expectation is for us to talk about subsequences--something not in the text at all. I wouldn't know how to justify it if my prof asked me $\endgroup$ Apr 14 at 0:19
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    $\begingroup$ You have dominated convergence, which is what I used in my replacement. Once you do the replacement monotone convergence comes into play. $\endgroup$ Apr 14 at 0:48
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    $\begingroup$ Have you been exposed to the sandwich theorem? If so, it's easy to apply it to your sequence because the general term $\ a_n\ $ satisfies $\ 0\le a_n\le\frac{2}{n}\ .$ $\endgroup$ Apr 14 at 10:09
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    $\begingroup$ Re: "The class is Calculus 2 so there is nothing of that technical rigor." Isn't the basis of any introductory calculus course the definition of a limit? What did you do in Calculus 1? How are you supposed to prove anything about a limit if you haven't defined it? $\endgroup$
    – Servaes
    Apr 14 at 11:49
  • $\begingroup$ @Servaes My memory of introductory calculus is a bit hazy now, but I'm sure there was a bit of hand waving over limits; analysis was when the hand waving stopped. $\endgroup$ Apr 16 at 0:16
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You could describe your sequence as follows, assuming it continues in the "obvious" way.

$a_n=\begin{cases}\frac{1}{m},\text{ if $n=2\cdot m-1$}\\ \frac{1}{m+2},\text{ if $n=2m$}\end{cases}$

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  • $\begingroup$ In our section of the textbook, there are no examples of breaking up the sequence into a piecewise function-esque sequence. Nor is there a mention of using subsequences (this is just a topic I'm familiar with). The class is Calculus 2. Are there any other tools we could use to prove convergence? The text has the Monotonic Convergence Theorem but the sequence is not decreasing for all n. And since I can't explicitly define it in a single function I'm not sure if the expectation is for us to talk about subsequences--something not in the text. I wouldn't know how to justify it if my prof asked me $\endgroup$ Apr 14 at 0:14
  • $\begingroup$ This notation is common. Maybe you can find a "closed form" for your sequence, but it is much simpler to state it like this. Keep in mind that my above answer has kind off nothing to do with your question. I just wanted to mention a way on how you could write down your sequence formally. To prove the convergence of the sequence, you would just follow the answer of Ross Millikan, and would basically perform a $\varepsilon$ proof. Since your subsequences converge to $0$, you will find an index such that every larger is less then $\varepsilon$, and use that information to conclude a proof. $\endgroup$
    – Cornman
    Apr 14 at 0:28
  • $\begingroup$ Cheers! Much appreciated. $\endgroup$ Apr 14 at 1:01
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    $\begingroup$ Please get in the habit of including a context in your questions. Like textbook the problem is from, course you're in, etc. $\endgroup$
    – nickalh
    Apr 14 at 9:45
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    $\begingroup$ For $\ n=2m\ $ the formula should be $\ \frac{1}{m+\color{red}{2}}\ .$ One closed form expression for $\ a_n\ $ is $\ \frac{4}{2n+5+3(-1)^n}\ ,$ but I agree that it's much simpler and easier to express it in the way you've done. $\endgroup$ Apr 14 at 9:55
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Am I missing something subtle here? It seems your original sequence is Cauchy, and you are working in a complete space, so the sequence converges.

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  • $\begingroup$ Yeah, this is a Calculus 2 student. It's likely he's never heard of a Cauchy sequence, which isn't usually introduced until Analysis. $\endgroup$
    – nickalh
    Apr 14 at 9:44
  • $\begingroup$ @nickalh Is Calculus 2 second year for US undergraduates? Here in New Zealand we started Analysis in 2nd year (or we did back in the 1960s). $\endgroup$ Apr 14 at 10:01
  • $\begingroup$ Oops, I forgot to specify country= USA. Calculus 2 for US undergrads, is usually taken in the freshman year. This is especially for students who passed the Advanced Placement AB Calculus exam for high school students. A few take the Advanced Placement BC course and exam, which covers Calculus 2 in high school. $\endgroup$
    – nickalh
    Apr 14 at 10:26
  • $\begingroup$ However, In the US universities,(maybe only my alma mater?)- University of Texas at Arlington) I'm aware of Analysis is a 4 semester sequence, comprised of 2 undergrad senior courses and 2 graduate senior courses. I think a few universities start the analysis sequence earlier, but am not familiar with those. Of course, with sufficient background, prerequisites, I think undergrads could take the Analysis courses as early as sophomore year. I've heard of Calculus of Variations and "Advanced Calculus" but am not familiar with them. $\endgroup$
    – nickalh
    Apr 14 at 10:30
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    $\begingroup$ Showing that the sequence is Cauchy is more difficult than just showing directly that it converges. $\endgroup$ Apr 14 at 16:27

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