Convergence or divergence of $\sum_{n=1}^\infty (-1)^{n+1}(2+(-1)^n)/n$?

Question

Is $$\sum_{n=1}^\infty (-1)^{n+1}{(2+(-1)^n)\over n}$$ convergent or divergent?

Answer 1

Hint: the sum of an even $n=2k$ term and the following term is $$-{3\over n}+{1\over n+1}=-{2n+3\over n(n+1)}=-{4k+3\over 2k(2k+1)}.$$
That looks divergent to me based on $\sum 1/k$ diverging.

Answer 2

Here I provide an answer using basic convergence tests as well as one without such tests

$$\sum_{n=1}^\infty (-1)^{n+1}{(2+(-1)^n)\over n}$$ $$=\sum_{n=1}^\infty (-1)^{n+1}\frac{2}{n} + \sum_{n=1}^\infty (-1)^{n+1}\frac{(-1)^n}{n}$$ $$=\sum_{n=1}^\infty (-1)^{n+1}\frac{2}{n} - \sum_{n=1}^\infty \frac{1}{n}$$
The first series is clearly convergent by the alternating series test, though the latter is the harmonic series, which is divergent. As a result, the original sum must diverge.

As an addendum, the first sum can be shown to be $\log(4)$ using Taylor Series, which is clearly convergent. Let's do so.

We know that the definition of the Taylor Series centered at $x=a$ is $\sum_{k=0}^{\infty} \frac{f^{(n)}(a)(x-a)^n}{n!}$. Let's use the fairly well known fact that $$\frac{1}{1-x} = 1+x+x^2+x^3\cdots$$
We then know that $$-\log(1-x) = \int_0^x \frac{1}{1-t}$$
And we combine these facts (assuming a lot of little things in the process) yielding $$-\log(1-x) = -\int_0^x \frac{1}{1-t} dt = \int_0^x \left(1+t+t^2+t^3\cdots\right) dt = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots = \sum_{x=1}^{\infty} \frac{x^k}{k}$$
When we evaluate this series at $x=-1$ we get
$$-\log(2) = \sum_{x=1}^{\infty} \frac{(-1)^k}{k}$$
Multiplying by $-2$ we get that $$2\log(2) = -2\sum_{x=1}^{\infty} \frac{(-1)^k}{k} = \sum_{x=1}^{\infty} (-1)^{k+1}\frac{2}{k}$$
Thus we have the series we desired.

If you want to show that the latter sum diverges without using tests then observe that the harmonic series is the following: $$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \cdots$$
The harmonic series will thus be greater than the following series at any finite point, as each term is larger $$\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\cdots$$
Notice, however, that the fractions form blocks, namely
$$\frac{1}{2}(1)+\frac{1}{4}(2)+\frac{1}{8}(4) + \frac{1}{16}(8) +\cdots$$
We can rewrite this as $$\frac{1}{2}+\frac{1}{2}+\frac{1}{2} + \frac{1}{2} +\cdots$$ And the above series clearly diverges. Since the harmonic series is greater at each finite point, the harmonic series must likewise diverge.