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?



2 Answers 2


To use the alternating series test, you need to rewrite the series first; you cannot use it directly with the $a_n$ you are given.

Define $b_n$ as follows: when $n$ is even, $b_n = \frac{3}{n}$; when $n$ is odd, $b_n = \frac{1}{n}$. Then $(-1)^nb_n = a_n$, so you can apply the alternating test to your original series by writing the series as $\sum (-1)^n b_n$.

Unfortunately, the alternating series test does not apply: remember that you need not only that $\lim\limits_{n\to\infty} b_n=0$, but also that $b_{n+1}\lt b_n$ for all sufficiently large $n$. However, for $n$ odd, we have $b_n\lt b_{n+1}$, since $\frac{1}{n}\lt \frac{3}{n+1}$ if and only if $n+1\lt 3n$, which holds for all $n$. So the alternating series test cannot be used.

The Ratio Test is likewise inconclusive: the limit does not exist. To see this, note that for $n$ odd, we have $$\left|\frac{a_{n+1}}{a_n}\right| = \frac{\quad\frac{3}{n+1}\quad}{\frac{1}{n}} = \frac{3n}{n+1}$$ which converges to $3$ as $n\to\infty$; whereas for $n$ even, we have $$\left|\frac{a_{n+1}}{a_n}\right| = \frac{\quad\frac{1}{n+1}\quad}{\frac{3}{n}} = \frac{n}{3(n+1)}$$ which converges to $\frac{1}{3}$. Since the sequence of odd terms and the sequence of even terms of $|\frac{a_{n+1}}{a_n}|$ converge to different things, $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$$ does not exist. So the Ratio Test is inconclusive.

As to your first method, it's fine: indeed, the sum of a divergent series and a convergent series must diverge. To see this, note that if $\sum r_n$ and $\sum(r_n+s_n)$ both converge, then so does $\sum ((r_n+s_n)-r_n) = \sum s_n$, since the difference of convergent series converges. Thus, if any two of $\sum r_n$, $\sum s_n$, and $\sum (r_n+s_n)$ converge, then so does the third, and in particular if any one of the series diverges, then at least two must diverge.


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.


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