The questions is:
Let $f$ be a function defined as $f(x) = (-1)^n/n $ for $x \in [n, n+1), n \in \mathbb{N}.$
Show that $lim_{n\to\infty}$ $\int_{[1,n]}\ f $ exists
Also, is $f$ integrable on $[1,\infty) $
Now, this function seems to closely mirror a sequence of simple (or step) function, however I do not know if it can be written as such and if that is even worth exploring. Any assistance on where to begin would be helpful. Please feel free to ask for any further clarification as well.
It seems to me that since this is a step function (piecewise constant, with each piece extending between integers) this is better written as a sum:
$$ \int_{x=1}^\infty f(x) \, dx = \sum_{n=1}^\infty \frac{(-1)^n}{n} = -\ln 2 $$
ETA: I suppose I had better explain where that last step comes from. One can write, for $z \not= 1$,
$$ \frac{1}{1-z} = 1 + z + z^2 + z^3 + \cdots $$
This is somewhat dicey, but if we take the (indefinite) integral of both sides, we get
$$ -\ln (1-z) = C + z + \frac{z^2}{2} + \frac{z^3}{3} + \frac{z^4}{4} + \cdots = C + \sum_{n=1}^\infty \frac{z^n}{n} $$
and by inspection at $z = 0$ we can quickly determine $C = 0$. The expression on the right converges absolutely only on the open interval $z \in (-1, 1)$, but it does converge conditionally at $z = -1$, so we write (with some trepidation!)
$$ -\ln 2 = -\ln (1-(-1)) = \sum_{n=1}^\infty \frac{(-1)^n}{n} $$
as desired. The general expression for $z$ is the Mercator series (q.v.).