Ytukyg

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.

The even positive integers can be partitioned into consecutive blocks $E_1 < E_2 < E_3 < cdots $ such that

$$sum_{nin E_k} 1/n > 2,,,text{for each } k.$$

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

$$sum_{n=1}^{infty} (-1)^{f(n)}frac{1}{f(n)} ge 1+1+cdots = infty.$$

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 $$sum_{ngeq 1} frac{(-1)^n}{n} $$ 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 $i,jgeq 1$, 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.

My answer addresses to the original question.

It's impossible to find a bijection $f: Bbb{N}toBbb{N}$ 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 $Bbb{N}$.