checking convergence of $\sum_{n=0}^{\infty}(-1)^{n}\frac{1-2n}{1-3n}$ 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!
 A: 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.
A: Recall that if $\sum a_n$ is a convergent series, then $a_n\to 0$. This is a consequence of the Cauchy criterion of convergence.
The $n$th term $(-1)^n\frac{1-2n}{1-3n}$ of the series doesn't converge to zero, so no more needs to be said. The series is divergent.
In fact, the analysis in your post shows that $a_n$ doesn't converge at all, let alone converge to zero.
A: Notice that it is the sum of $2$ series that don't converge since the terms never go to zero. Here's what I mean. Write it like
$$\sum_{k=0}^{\infty} \left(\frac{1-2(2k)}{1-3(2k)}-\frac{1-2(2k+1)}{1-3(2k+1)}\right)=-\frac{1}{2}\sum_k \frac{1}{ (3 k + 1) (6 k - 1)}\approx 0.457​85$$ and use the new $a_k$ for your ratio test, or whatever test you like. My favorite is the Gauss' Test.
A: If we can show absolute convergence for your series, then the original series converges. That is, if $\sum_{n = 0}^{\infty}{|a_n|}$ converges, $\implies \sum_{n = 0}^{\infty}{(-1)^n a_n}$ convergence.
Given, $|a_n| = \frac{2n-1}{3n-1}, n \ge 0$, we can prove inverse implication of convergence by $\sum_{n = 0}^{\infty}{|a_n|}$ convergence $\implies \prod_{n=0}^{\infty}(1 + |a_n|)$ convergence.
$$\prod_{n = 0}^{k}{1 + \frac{2n - 1}{3n - 1}} = \prod_{n = 0}^{k}{\frac{5n - 2}{3n - 1}} = \frac{5\big(\frac{5}{3}\big)^k\big(k - \frac{2}{5}\big)!\big(\frac{-4}{3}\big)!}{3\big(k - \frac{1}{3}\big)!\big(\frac{-7}{5}\big)!}$$
However, $$\lim_{k \to \infty}{\left( \frac{5\big(\frac{5}{3}\big)^k\big(k - \frac{2}{5}\big)!\big(\frac{-4}{3}\big)!}{3\big(k - \frac{1}{3}\big)!\big(\frac{-7}{5}\big)!} \right)} \to \infty\cdot0  \text{  (indeterminate)} $$
which suggests divergence, and hence $\sum_{n = 0}^{\infty}{|a_n|}$ diverges too.
Futhermore, $$\lim_{k \to \infty}\left(\sum_{n = 0}^{k}{\ln{\left( \frac{2n - 1}{3n - 1} \right)}}\right) = \lim_{k \to \infty}\left(\ln{\prod_{n = 0}^{k}{\left(\frac{2n - 1}{3n - 1}\right)}}\right) \to \ln{0}$$ Since the inside product converges to $0$, the logarithmic summation diverges. $\frac{2n - 1}{3n - 1} > \ln{\left(\frac{2n - 1}{3n - 1}\right)}, n > 0$ and thus (through absolute convergence stated earlier), your summation diverges too.
A: Using Lerch transcendent function
$$S_p=\sum_{n=0}^{p}(-1)^{n}\frac{1-2n}{1-3n}$$
$$S_p=-\frac{9 +3 \log (2)-\pi\sqrt{3}  }{27}-\frac{(-1)^p }{9}  \left(\Phi\left(-1,1,\frac{3p+2}{3}\right)-3\right)$$
For large values of $p$, $\Phi\left(-1,1,\frac{3p+2}{3}\right) \to 0$ which makes
$$S_p \sim -\frac{9 +3 \log (2)-\pi\sqrt{3}  }{27}+\frac{(-1)^p }{3}$$
