Show that the telescoping series below converges if and only if the
$\lim_{j\to\infty} c_j$ is defined and finite.$$\sum_{j=1}^{\infty} c_j – c_{j+1}$$
Not really sure where to start for this, proofs are nowhere near my strong suit. Would $c_j$ not be a constant? Why would I take the limit of a constant?
Gotta go to class now, will check back this afternoon. Thanks in advance.
Edit: my progress (from a reply below) that anyone can comment on:
I'm currently trying to look at this and still not seeing it. Here is what I've played with. As you work through the sum (Starting at 1, I typo'd in the original question) you have: $(c_1-c_2)+(c_2-c_3)+(c_3-c_4)+…(c_{n-1}-c_n)+(c_n-c_{n+1})$.
For every term other than the first and last it gets subtracted and then added, essentially giving a sum of $c_1-c_{n+1}$. the last term is going to be either substantially bigger or substantially smaller than c1, but I'm still missing something huge if I'm even on the right track. Am I even on the right track?
Edit2: Here is my rough answer as it currently stands, is it missing anything?
A rough "proof-ish" description the answer as I think I have it now: Because of the telescoping nature of the series, every term after the first and except for the last is cancelled out by the one after it. This leaves us with a partial sum of Sn=c1-cn+1. Because c1 is finite, in order for the sum to converge lim(cn+1) cannot be infinite and must be defined.
Best Answer
For a fixed $j$, $c_j$ is a constant, but if we let $j$ change, then $c_j$ is not a fixed constant (for example, we may consider $c_j=1/j$). Now, my hint will be to approach the problem the following way: $$\sum_{j=0}^\infty c_j-c_{j+1}=\lim_{N\to\infty}\sum_{j=0}^N c_j-c_{j+1}.$$Now the sum on the right-hand side is finite, so what happens if we write out a few terms? I'd suggest trying it when $N=2$ or $N=3$ and deduce what happens for all larger $N$.