What is the Convergence of the Series in Rudin's Exercise?

[TPAINE]

Homework Statement

Prove 1+1/3-1/2+1/5+1/7-1/4+1/9+1/11-1/6... converges.

3. Relevant Info

This was left as an exercise to the reader in the text of Rudin. Both the root and ratio tests fail, and I don't see anything obvious. I found a proof online that it converges to 3/2 log 2, but it is quite involved. I am looking for a quick way to prove it converges; I do not care about finding the exact value.

 

[henry_m]

With a slight modification, you can use the alternating series test.

 

[TPAINE]

I'm assuming you are proposing grouping the positive terms and using the alternating series test.

I'm not sure how I would rigorously justify using that. That would would show that, for positive integer n, the partial sums of the first 3n and 3n+2 terms converge, but it says nothing about the partial sums of the sequence when considering the first 3n+1 terms, for any $n$. I think it is conceivable that for another sum, the terms that are 1 mod 3 could be "weird" enough that they cause the sequence to diverge, even though the partial sums of the k terms, where k is 0 or 2 mod 3, converge.

 

[TPAINE]

For example, consider 1-1+1/2+1-1-1/3+1-1+1/4+1-1-1/5... This diverges, because if it had a limit L, choose epsilon to be 1/3. Then at some point, the alternating 1,-1s are going to make the sequence be more than 1/3 away from L. But if we try to naively group them, as you propose above, we get the standard 1+1/2-1/3+1/4-1/5... which converges.

 

[henry_m]

I see your concern, and you're right that we should be careful. But it shouldn't be a problem, as long as you only group together positive terms or negative terms, and not both. To be explicit, here's a statement of a generalisation of the alternating series test that will do the job:

Suppose we have a sequence a_n, with infinitely many positive and negative terms. Let a 'group' be a maximal collection of consecutive terms of the same sign (let's put 0 with the +'s for definiteness). So the sequence consists of a positive group followed by a negative group followed by a positive group etc. Let the 'value' of a group be the absolute value of the sum of its terms. Then if the sequence of group values is decreasing and tends to zero, the series converges. The alternating series test is the special case where the groups all contain one term.

The proof is essentially the same as for the special case. Let's introduce some more language: let a 'maximum' be a partial sum up to the last term of a positive group, and a 'minimum' be a partial sum up to the last term of a negative group. Now, since the value of any group is smaller than the value of the one that preceded it, the sequence of maxima is decreasing, and each maximum is an upper bound for all subsequent partial sums. Similarly, the minima form an increasing sequence of lower bounds. Finally, since the sequence of values tends to zero, the sequence of maxima and the sequence of minima converge to a common value. Then the series itself must also converge to this value.

In the example you give, where you group 1s and -1s together, these would be split into separate groups and so the values of the group would not tend to zero. OTOH, the original sequence does satisfy the conditions of the test.

 

[TPAINE]

Excellent exposition. Thank you. To be clear, in your third paragraph you are using a limsup/liminf argument?

 

[henry_m]

Excellent exposition. Thank you. To be clear, in your third paragraph you are using a limsup/liminf argument?

Yes, you could certainly phrase it in that language if you like. Glad to be of some help!