I'm trying to test endpoints for integral of convergence, and this one has been giving me a fit…
$$\sum_{n=0}^{\infty}(-1)^{n}\frac{1-2n}{1-3n}$$
I tried the alt. series test, which was inconclusive, I tried the ratio test, which was inconclusive, I tried absolute convergence, which was no help since it did not converge absolutlely…. then I had the thought $(-1)^{n}=1\vee -1$… so, couldn't I just test each condition individually using the regular diversion test?
Doing so I get:
$$\lim_{n \to \infty}\frac{1+2n}{1+3n}=\frac{2}{3}\neq 0\rightarrow \text{divergent}$$
and
$$\lim_{n \to \infty}-\frac{1+2n}{1+3n}\rightarrow -\lim_{n \to \infty}\frac{1+2n}{1+3n}=-\frac{2}{3}\neq 0\rightarrow \text{divergent}$$
Is there anything wrong with this approach? It seems that it may be inappropriate since the conversion/diversion of $a_n$ and $-a_n$ may not correlate to an alternating series. If I'm off-base here, would someone mind pointing me in the right direction?
Thanks!
Best Answer
Your approach to alternating series convergence is correct and avoids the trap/trick in the question.
A series is convergent if its partial sums have a limit. If some terms don't approach 0, the partial sum will change drastically at those terms, and so, the partial sums won't converge.
The trap is that you can have series where a subset of partial sums converge. Eg if you take $b_n=a_{2n}+a_{2n+1}$, you can get a convergent series.
I believe that's the case here, and you're correct in not pursuing that avenue.