How do i show my decreasing sequence converges to its infimum?

[Chloe_S]

I'm working through a problem about the convergence of a sequence defined by a recurrence relation, and I've hit a wall trying to apply the monotone convergence theorem. My sequence is decreasing and bounded below, so I know it should converge, but I'm stuck on formally proving the bound is the infimum.

[Sofia_L]

Take L as the limit of the decreasing sequence. Since it’s bounded below, the limit exists. That makes L a lower bound because a_n ≥ L for every n. To pin down the infimum, use the convergence: for any ε > 0 there is an N with a_N < L + ε. If you try any number b > L as a lower bound, pick ε = b - L > 0; then a_N < b, so b can’t be a lower bound. So the usual ε–N argument lines up with the bound being the infimum.

[Edward.M]

I remember the moment I tried this, I wrote down a few terms and checked how far they were from the suspected limit. I tracked a_N − L and watched it shrink from something like 0.4 to 0.05 as N grew. That concrete check helped me see how the ε-quantifier plays out, even if I hadn’t written a full proof yet.

[MatthewS]

Maybe the real snag isn’t the logic so much as whether L you get is the bound you actually care about. I once worried I had the wrong bound because I assumed monotone convergence auto gives the infimum, and then I realized I needed that ε-subinterval argument too. It felt flaky until I forced myself to pick ε and find N. While I was thinking about that, I started wondering if a different part of the recurrence could blow up, but it didn’t.

[JonathanH]

A compact way to test it: assume a_n decreases to L. Then for every ε>0 pick N with a_N < L + ε; since a_n ≥ L, no number bigger than L can be a lower bound. That’s the core idea, even if you’re wary about the last step.