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Assuming the first 5 terms of a sequence

$$\frac{5}{3},\frac{4}{3},\frac{11}{12},\frac{7}{12},\frac{17}{48}$$

Find the general term.

There doesn't seem to be a pattern here. Correct me if I'm wrong, but I would say there is no general term.

If there is no general term ($n$th term), is it safe to say that the sequence diverges by default?

Eric Johnson
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  • Not at all---"seemingly has no general term" is not well defined, and in some sense /most/ convergent sequences "seemingly have no general term". – Travis Willse Apr 21 '16 at 18:08
  • It is not sufficient. You could still have a "unpredictable" sequence approach a constant value. Take for example digits of $\pi$ over increasing integers, this would approach zero even though I can't tell you a formula for the $n$th term. – Carser Apr 21 '16 at 18:09
  • @Travis Does the sequence above has a general term? If not I will remove "seemingly". If it does, I will remove this sequence from the question – Eric Johnson Apr 21 '16 at 18:09
  • Your sequence is positive and decreasing, at least so far – grand_chat Apr 21 '16 at 18:10
  • It is possible to fit a polynomial from your 5 terms, do you accept that as a generating formula? – peterwhy Apr 21 '16 at 18:11
  • Having a sequence means exactly that we have every term in the sequence, i.e., a general term. (This is not the same thing as having, e.g., a simple closed-form expression for the term.) On the other hand, the question doesn't actually specify a sequence, it only gives a few terms of one. – Travis Willse Apr 21 '16 at 18:12
  • Someone needs to tell me whether the the above 5 first terms of a sequence have a general term so I can appropriately edit my question – Eric Johnson Apr 21 '16 at 18:12
  • @EricJohnson If you give us a finite number of real numbers and ask us to find a sequence using those numbers as the beginning terms, there are literally an infinite number of sequences or general formulas that would fit those terms because we could make a formula where the next term could literally be any other real number using polynomials and common differences. It's really hard to do this and I forgot the name of the theorem that lets us do this, but I've done it before many times. – Noble Mushtak Apr 21 '16 at 18:13
  • @EricJohnson As I just wrote, the question of "whether there's a general term" of the sequence is meaningless, as the question doesn't actually specify a sequence. – Travis Willse Apr 21 '16 at 18:15
  • @peterwhy Surely not, since OP is asking about convergence. – Travis Willse Apr 21 '16 at 18:16
  • @peterwhy no it's not, I actually specified "5 first terms of a sequence" – Eric Johnson Apr 21 '16 at 18:16
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    Note that the sequence so far is $5/3, 8/6,11/12,14/24,17/48$. It is not unreasonable to think that the next term is $20/96$ (add $3$ to the previous top, double the previous bottom). One could simplify this to $5/24$, but that would hide the pattern. Remark: When you are trying to detect a pattern, "simplifying" is often a very bad idea, and calculating the numbers with a calculator to $7$ decimal places is an even worse idea. Calculators produce a jumble of digits, and ordinarily make it much harder to detect a pattern. – André Nicolas Apr 21 '16 at 18:21
  • @EricJohnson, by inspection, the following general term (piecewise definition) seems to work.

    $$a_n=\begin{cases}\dfrac{3n+2}{3\times 2^{n-1}}~\forall~n\equiv 1\pmod 2\ \dfrac{3n-2^{n/2}}{3\times 2^{n-2}}~\forall~n\equiv 0\pmod 2\end{cases}$$

     – learner Apr 21 '16 at 18:26
  • @AndréNicolas how did you detect the pattern ? – Eric Johnson Apr 21 '16 at 18:28
  • @learner same question? – Eric Johnson Apr 21 '16 at 18:28
  • @EricJohnson: Just looked. The second and fourth term had the "wrong" denominator. – André Nicolas Apr 21 '16 at 18:30
  • @EricJohnson: If we assume the pattern we detected continues (and it need not) the limit is $0$, for the bottom grows exponentially while the top only grows linearly. – André Nicolas Apr 21 '16 at 18:33
  • @AndréNicolas brilliant, if you care to write it up i'll hand it to ya – Eric Johnson Apr 21 '16 at 18:35
  • @peterwhy had a similar answer, which I hope gets undeleted. – André Nicolas Apr 21 '16 at 18:37
  • @AndréNicolas your call – Eric Johnson Apr 21 '16 at 18:38

1 Answers1

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If there is no definition for $a_n$ for all $n \in \Bbb{N}$, then it's not a sequence. There needs to be a term for each natural number $n$ for it to be a sequence.

Sequences don't have to have a particular pattern like arithmetic or geometric. They simply have to have some sort of term mapping to each $n \in \Bbb{N}$. I don't notice any particular pattern with your sequence other than it's decreasing, but if it keeps decreasing and yet stays positive, it could certainly converge to some value like $0$. However, if it keeps decreasing and goes down toward negative infinity, it could certainly diverge. There's no way to tell what the pattern is if you just give us the first few terms.

The thing is, there are literally infinite number of general formulas for this sequence because you only gave us a finite number of terms. This is really hard to explain, but this answer seems to do a pretty good job.

Noble Mushtak
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