I can't figure out my mistake in calculating the harmonic series

Question

Let's say $a = H_{\infty}$, so $a = \sum_{k=1}^\infty \frac{1}{k}$. $$a = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{\infty}$$ Now if we take $a - a + 1$ we get $$a - a + 1 = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{\infty} - (1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{\infty}) + 1 $$ $$=1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{\infty} - \frac{1}{2} - \frac{1}{3} - \cdots - \frac{1}{\infty} = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{6}) + \cdots$$ $$=\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ... + \frac{1}{\infty}$$ That means that $$a - a + 1 = 1 = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ... + \frac{1}{\infty} = \frac{1}{2}a$$ $$1 = \frac{1}{2}a$$ $$a = 2 \cdot 1 = 2$$ Now since I now the harmonic series are divergent, I know the result I got is false. However, I tried to used the same trick that's used to proof that the Grandi's series ($\sum_{n=0}^{\infty} (-1)^n = 1 - 1 + 1 - 1...)$ are equal to $\frac{1}{2}$. My only question is, where did I go wrong?

Answer 1

All the answers are given in the comments, so if the users who posted these comments could post it as answer, this answer will be deleted and their answer will be chosen as accepted answer. Until then, here is a summary. To sum it up (padum tss):

The main mistake made is assuming that $\frac{1}{2}$ is the answer to Grandi's series ($\sum_{n=0}^{\infty} (-1)^n$), which it's not (the Grandi's series are divergent). Because of that, we can't use the same trick, since it will not result in the right answer. $a$ is not a real number, and therefore should not be treated like one. That basically is the mistake I made.

EDT: What I also just found out is that I don't subtract the $\frac{1}{3}, \frac{1}{5}, \frac{1}{7}$ and all other odd fractions, so $$a - a + 1 = \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + ... \frac{1}{\infty}$$ which of course limits to $0$. That makes quite a bit more sense, even though, as stated above, is wrong.

Answer 2

You can't do the step on line 5, because of https://en.wikipedia.org/wiki/Riemann_series_theorem.