Cauchy critereon question.

This is a very convenient tool in many intro analysis proofs (and just proofs in general). By adding and subtracting L, you've done nothing. The L's will cancel so the value inside the absolute value brackets hasn't been changed at all. I didn't watch any further, but I'm assuming he then uses the triangle inequality to break that expression up into two terms, which is a pretty standard "trick".

an - am = (an-L) - (am-L)

And then triangle inequality says

|an - am| ≤ |an-L| + |-(am-L)|

And |-(am-L)| = |am-L| so

|an - am| ≤ |an-L| + |am-L|
Really there are two possible questions you might have. Why is an-am=(an-L)-(am-L)? And why is it useful to do this seemingly convoluted step?

As for the truth of the equality: since L-L=0, and adding 0 doesn't change anything, we have an-am=an-am+L-L. Now just group an-L and -am+L together to obtain that this is equal to (an-L)+(-am+L), which in turn is equal to (an-L)-(am-L).

Now for why this is useful: convergence of a sequence to L allows us to place neat bounds on terms of the form |an-L|. A sequence being Cauchy, on the other hand, means that terms of the form |an-am| can be bounded in a certain way. Now in this proof, we *know* that we can bound |an_L| and |am-L|, and we *want* to bound |an-am|. So our best bet is to bound |an-am| by the two terms we *can* bound. In order to do this, we need to introduce two copies of L, as is done above.