Is this sequence convergent or divergent?

[cue928]

For the following sequence of numbers {1, 1/3, 1/2, 1/4, 1/3, 1/5, 1/4, 1/6,...} determine if it is convergent or divergent.

Since several numbers appear more than once, my initial thought was that it did not converge. But, the numbers are getting smaller. So I thought to proceed with coming up with a formula for the denominator but was unable to find one. I've done this on several others with no problem but did not see it on this one.

 

[micromass]

Hi cue928,

How rigourous does your proof need to be?

Firstly, do you see what the denominators do in this sequence? It always goes +2, then -1, then +2, then -1,...

Thus, you start with 1
+2, then you get 3
-1, then you get 2
+2, then you get 4
-1, then you get 3

Your guess was indeed correct, this sequence keeps on getting smaller, and hence converges to 0. But again, how rigourous does this need to be?

 

[cue928]

See I don't know. For example, one of the problems was an=n^3 / (n+1). It sufficed there to divide thru by n, take the limit, and we're done. So I don't know if they even wanted a proof in the true sense of the word. My guess is that I need to provide something more than intuition.

Is there a formula that would yield that sequence of numbers? I tried but could not find it. I did have the +2, -1...differences that you showed.

 

[Maxter]

Note that this is a sequence (terms listed with commas), and not a series (terms added).

You can come up with a formula, but you have to break it into even and odd terms, and you can get a formula in terms of n if you look at the odd terms a_{2n+1} and the even terms a_{2n} separately. That might be more than you are asked to do though - have you seen things like that in class?

 

[micromass]

Well, you could start by noticing that the odd indices form a sequence 1/n. Thus x_{2n-1}=1/n for n starting with 1.
The even indices also behave as 1/n, but a bit shifted, thus x_{2n}=1/(n+2), for n starting with 1.

This gives you a nice formula...