Let . Since the series converges absolutely, the sequence converges (and is Cauchy)[/tex]. Rewriting the series you are given:
Writing out a few elements of this series, you have
Collecting like terms, we have
So, the general form of the series seems to be . Use induction to prove that this is true, and then use the Alternating Series Test to prove that it converges conditionally.
I am not quite sure how to apply that, but I thought of something else instead that might be helpful. Isn't a subsequence of which is convergent so any subsequence would also be convergent?
Or because is a partial sum of a convergent series , also converges.
Then we would just need to apply . Not sure how to do that?
A subsequence is very different from a partial sum. A sum of partial sums may not converge. But, I think I figured it out.
Now, can you show that for almost all ?
Since converges and for all n, it must be that . So what I am getting at,
shows up in the first term only. shows up in the second term only. shows up in terms 2 and 3. shows up in terms 3 and 4. shows up in terms 3, 4, and 5. shows up in terms 4, 5, and 6. Seeing a pattern?
Also, this expansion should show you how each term is a sum of n elements over n.