TITLE: Offset Alternating Series
QUESTION [7 upvotes]: I have the following alternating series that I would like to determine whether it is absolutely convergent, conditionally convergent, or divergent:
$$  \sum\limits^{\infty}_{n=1} \frac{1+2(-1)^n}{n} $$
I have applied some tests and I find it reasonable to conclude that it is divergent.
As a sum of two series:
$$  \sum\limits^{\infty}_{n=1} \frac{1}{n} + \sum\limits^{\infty}_{n=1} \frac{2(-1)^n}{n} $$
I believe a convergent series when added to a divergent series, results in a divergent series.  If this isn't a fact then I would still be left to say that it is inconclusive.
Using the Alternating Series Test, with:
$$ a_n = \frac{1+2(-1)^n}{n} $$
although this isn't of 'proper form' $  \sum\limits^{\infty}_{n=1} (-1)^n a_n $ the limit of $a_n $ does approach zero as $ n \rightarrow \infty $.  As for monotonically decreasing, the limit of the ratio of absolute terms is divergent for $ n $ even and inconclusive for $ n $ odd, which has me concluding divergent by The Ratio Test as well as not monotonically decreasing, where:
$$ \lim\limits_{n \rightarrow \infty} \left\lvert \frac{2(-1)^n + 1}{2(-1)^n - 1} \frac{n}{n+1} \right\rvert $$
Am I on the right track here?  Am I making any really improper assumptions?  Was there a better way to go about with the proof?
Thanks!

REPLY [2 votes]: If we pair the terms with 2n and 2n+1, we get,
using Arturo Magidin's notation,
$$b_{2n}-b_{2n+1} = \frac{3}{2n} - \frac{1}{2n+1}
=\frac{3(2n+1)-2n}{2n(2n+1)}
= \frac{4n+3}{2n(2n+1)}
> \frac{1}{n},
$$
so the sum of the first 2n+1 terms
is greater than 
$\sum_{k=1}^n \frac{1}{k}
$
which diverges.
In general, when working with an alternating series.
I find it useful to pair each even term with the following odd term.
This can be readily generalized to the case where
the signs of the terms of a series to be summed
follow a repeating pattern.