I'm currently reading Vector Calculus, Linear Algebra, and Differential Forms (Hubbard and Hubbard) and have reached the definition of convergence (Chapter 1.5, Page 76 in my edition). Most of it makes sense, but I am confused by one element of it. Below is the definition.
Definition 1.5.8 (Convergent sequence). A sequence of points in $\mathbb{R}^n$ converges to $\boldsymbol{a} \in \mathbb{R}^n$ if for all $\epsilon > 0$ there exists $M$ such that when $m > M$, then $|\boldsymbol{a}_m-\boldsymbol{a}| < \epsilon$. We then call $\boldsymbol{a}$ the limit of the sequence.
My confusion is why one needs $M$, since all it is used for is to find a number larger than it. What is the difference between the above and the statement without the $M$: [It converges] if for all $\epsilon > 0$ there exists $m$ such that $|a_m-a| < \epsilon$?
The problem with your phrasing is that you just require there to be a single value of $m$ for which the sequence is within $\epsilon$ of $a$. That's easily satisfied by sequences that don't approach the limit appropriately - for example, consider the sequence $a_m = m$.
I claim that $a = \lim_{m \rightarrow \infty} a_m = 3$. For any $\epsilon > 0$, take $m = 3$. Clearly $|a_3 - a| = |3 - 3| = 0 < \epsilon$, so we're all good.
The idea of having $M$ in there is that you need to be able to say that for a given $\epsilon$, then all the terms $a_M, a_{M+1}, \ldots$ are within $\epsilon$ of $a$ - essentially, you can always find an $M$ such that after you cut off the first $M - 1$ terms you've only got numbers in that neighbourhood.