Series $\sum(-1)^{a(n)}\frac1{a(n)}$ converging to infinity with $a:\mathbb N\to\mathbb N$ a bijection I'd love to get a hint or direction on this and figure out a solution myself:
Construct a bijection: $$\: f:\mathbb{N} \rightarrow \mathbb{N} $$
such that:
$$ \sum_{n=0}^{\infty} -1^{f(n)}\frac{1}{f(n)} = \infty $$
I thought of something like below to eliminate all series elements that are negative:
$$
f(n):= 
\left \{
\begin{array}{ll}
n^{odd} \rightarrow n-1 \\
n^{even} \rightarrow n\\
\end{array}
\right.
$$
But then $f(n)$ cannot be a bijection. Another way would be to transform the fraction $\frac{1}{f(n)}$ into something that is divergent, but I don't see how that would be possible in an $\mathbb{N} \rightarrow \mathbb{N}$ map.

EDIT: Deleted the suggestion that the resulting series would be a harmonic or geometric series.

 A: The even positive integers can be partitioned into consecutive blocks $E_1 < E_2 < E_3 < \cdots $ such that
$$\sum_{n\in E_k} 1/n > 2\,\,\,\text{for each } k.$$
Now think of listing $\mathbb N$ as $E_1$ followed by $1,$ then $E_2$ followed by $3,$ then $E_3$ followed by $5,$ etc.  This defines a bijection $f$ of $\mathbb N$ onto $\mathbb N$ for which
$$\sum_{n=1}^{\infty} (-1)^{f(n)}\frac{1}{f(n)} \ge 1+1+\cdots = \infty.$$
A: I don't think you mean a geometric series because that's not what you've written so my answer will be pointing you towards a rearrangement of the series $$\sum_{n\geq 1} \frac{(-1)^n}{n} $$ which will diverge. We can do this because that series above is conditionally convergent. 
Let $a_i = \frac{1}{2i}$ and $b_j = -\frac{1}{2j-1}$ for $i,j\geq 1$, then we will arrange the $a_i$ and $b_j$ so that their sum converge. Let $n_1$ be minimal such that $$a_1 + a_2 + \cdots + a_{n_1} \geq -b_1 + 1$$ Now let $n_2$ be minimal such that $$a_{n_1+1} + \cdots + a_{n_2}\geq  -b_2 + 1$$
and continue in this way. Then $a_{n_{k-1}+1} + \cdots+ a_{n_k} + b_k \geq 1$ for all $k$  and so $$a_1 + a_2 + \cdots + a_{n_1} + b_1 + a_{n_1 + 1}  + \cdots + a_{n_2} + b_2 + \cdots $$ diverges. Hopefully this should be enough for you to construct your bijection.
A: My answer addresses to the original question.
It's impossible to find a bijection $f: \Bbb{N}\to\Bbb{N}$ such that the alternating sum of the reciprocals is an alternating divergent geometric series.
To make $$\sum_{n=0}^{\infty} (-1)^{f(n)}\frac{1}{f(n)} = \infty$$ an alternating divergent geometric series, you need $\dfrac{a_{n+1}}{a_n} = -M < -1$ for some constant $M > 1$, where $a_n = (-1)^{f(n)}\dfrac{1}{f(n)}$.  (You need constant ratio $\dfrac{a_{n+1}}{a_n}$ for geometric series, and if $0 < M < 1$, the series is absolutely convergent.  If $M = 1$, $f$ is not a bijection.)  The common ratio becomes $$-M = \frac{a_{n+1}}{a_n} = (-1)^{f(n+1) - f(n)}\frac{f(n)}{f(n+1)}.$$
Since the codomain of $f$ is $\Bbb{N}$, $\dfrac{f(n)}{f(n+1)} = M > 1$.  It's easy to see that $f(n) = \dfrac{1}{M^n} \in (0,1)$, contradicting the condition that the codomain of $f$ is $\Bbb{N}$.
