A Series Convergance Problem

I don't know why, but I have just overcome with the urge to write up a SLOG post about problem-solving, specifically a problem of mathematical nature. So let's waste no time, and get to it!

If you remember from my first post, one of my favorite courses from last year happened to be MAT137. And one of my favorite topics in that course had to do with proofs regarding the convergence/divergence of various series. As such, I decided to dig through my stuff from last year, and find a question that I didn't get around to attempting to solve last year. A lemma from the textbook caught my eye, whose proof was left to the students as an exercise:

Assume that some series (SIGMA)a_n is convergent, and that an => 0 for all n in N(natural numbers). Prove that the series (SIGMA)(a_n)^2 also converges.

Instead of chicken scratches on a piece of paper, I decided to take the organized approach Prof. Heap referred to in the SLOG handout.

1. Understanding the Problem
Let's think about what it means for a series to be convergent. Mathematically, this means: For all ε > 0, there exists there exists some N’ in N(natural numbers), so that for all n >= N’, |a_n| < ε. So knowing nothing about sequence “a_n,” I would have to somehow show that the same thing is true about sequence “b_n = (a_n)^2”. This condition certainly doesn’t seem impossible.   

2. Devising a Plan
This problem looks simple enough, but it is a bit different from the other series behaviour problems I've solved so far. From my recollection (which, I admit, is a bit wonky), either the sequence "a_n" on which the series is based on is usually explicitly given, or the formula for the series can either be: a) manipulated to a sum that has easily provable behaviour, or b) can be linked with a function whose end-behaviour can be proven using a variety of theorems. But in this case, I know nothing about the series or the sequence "a_n" on which the series is based on.
I have a few ideas about how to manipulate the relationship |a_n| < ε or - ε < a_n < ε, to get some facts about (a_n)^2, namely multiplying out the inequality by “a_n” throughout. Then, I believe, I could use the pinching theorem to justify the condition for “b_n = (a_n)^2”.

3. Carrying out the Plan
  Assuming a_n converges.
            Then |a_n| < ε.
            Then - ε < a_n < ε.
            Then – ε(a_n) < a_n(a_n) = ((a_n)^2) < ε(a_n).
            Then it suffices to show the two series bounding (a_n)^2 converge for (a_n)^2 to  
             converge.
            For – ε(a_n): – ε(– ε) > – εa_n > – ε( ε).
            For   ε(a_n): ε(– ε) <  εa_n <  ε( ε).
            So (a_n)^2 is bound between – ε( ε) and ε( ε).
            So –ε^2 < - εa_n < (a_n)^2 < εa_n < ε^2. And by the  pinching theorem, (a_n)^2   
            converges.
Then, given that a_n converges, then (a_n)^2 must converge.

And there you go! It was that simple. I wish I did this last year, so I could totally feel like a mathematics prodigy.