In this problem, we are given the defintiion of e, a number you'll see a lot in precalculus and calculus. When we are evaluating, a₁₀, it means that 10 goes in for n. If it's a₂₂₂₂, then 2222 goes in for n. When n gets large enough, the approximation gets closer and closer. Only as we get closer and closer infinity are we closer and closer to the exact number e.
a₁₀ = (1 + (1/10)¹⁰ = 1.1¹⁰ = 2.5937
a₁₀₀ = (1 + (1/100)¹⁰⁰ = 1.01¹⁰⁰ = 2.7048
a₁₀₀₀ = 1.001¹⁰⁰⁰ = 2.7169
a₁₀₀₀₀ = 1.0001¹⁰⁰⁰⁰ = 2.71814
a₁₀₀₀₀₀ = 1.00001¹⁰⁰⁰⁰⁰ = 2.71826
As n gets larger, the approximation gets better.
