I know there are probably many questions about the sum $1+2+3+4+\cdots = -1/12$, but I am wondering about the method I used which lead me to this result which was not my intention.
I am lacking proofs for a lot of what I do, as this was just me trying to see what happens. Also, to cut down on the amount of latex, I will use the zeta function to replace the sum
I had decided to see if I would be able to evaluate the sum $\sum_{n\geq 1} 1/n^2$ and so my first attempt was to try and solve
$$ S(x) = \sum_{n\geq 1} \frac{x^n}{n^2}$$
but ended up with the integral
$$-\int_0^x \frac{\ln(1-x)}{x} \, dx$$
and was unable to evaluate it (I later learned this has no elementary antiderivative). So I thought that solving a sum whose terms are of the form $y(x)/n^2$, where I could pick some $x'$ such that $y(x') = 1$ might be solvable and eventually arrived at
$$\sum_{n \geq 1} \frac{\sin(n\theta)}{n} = \frac{\pi - \theta}{2}$$
Now I know very nearly nothing about analysis and so I took some liberties and just went ahead and integrated this to get
$$-\int_0^\theta \sum_{n \geq 1}\frac{\sin(n\omega)}{n}d\omega = \zeta(2)-\sum_{n\geq 1}\frac{\cos(n\theta)}{n^2} = -\frac{1}{4}\theta\,^2+\frac{\pi}{2} \theta$$
which results in $$\zeta(2) - \sum_{n \geq 1}\frac{(-1)^n}{n^2} = \frac{\pi^2}{4}$$
by letting $\theta = \pi$. This leads me to the answer of $\zeta(2) = \pi^2/6$.
This method can be continued to calculate any of the sums whose terms are of the form $1/n^{2k}$. I then thought about starting with $\cos(n\theta)/n$ just to see what happens. I ended up with
$$\sum_{n \geq 1} \frac{\cos(n\theta)}{n} = -\frac{1}{2}\ln(2-2\cos(\theta))$$ Though this cannot be integrated, by taking derivatives (again, lots of liberties and ignorance) I quickly arrive at many of those divergent sums such as $1-1+1-1+\cdots = 1/2$ and $\sum_{n \geq 1} n^{2k+1}$, as in the title I quickly arrived at $1+2+3+\cdots = -1/12$.
Is this just some complete fluke, or is there a reason that these pop up?
[edit] I wanted to add that I have a copy of introduction to analysis by Rudin and I am just starting to go through that so that i can give more rigorous proofs