Pointwise Convergence & Uniform Convergence

Question

How do we prove that f_n(x) =[1+(x/n)]ⁿ on [0,2] is uniformly convergent or not.

Answer 1

Consider the sequence g_n(x) := log f_n(x) on [0,2]. The mean value theorem shows log (1 + t) ≤ t for all t ≥ 0; it then follows that 0 ≤ g_n(x) ≤ x for all x ≥ 0. If x > 0, another application of the mean value theorem results in the equation x - g_n(x) = (1 - 1/(1 + c/n))x where 0 < c < x. Note

1 - 1/(1 + c/n) = c/n / (1 + c/n) < c/n < x/n

so that

0 ≤ x - g_n(x) < x² / n

Since x² / n ≤ 4 / n on [0, 2], it follows that |x - g_n(x)| ≤ 4/n for 0 ≤ x ≤ 2.

The mean value theorem applied to the function e^x over the interval [0,2] yields the inequality |e^x - e^y| ≤ e²|x - y| for all x, y in [0,2]. Hence, for every x in [0,2],

|e^x - f_n(x)| = |exp(x) - exp(g_n(x))| ≤ e² |x - g_n(x)| ≤ 4e² / n

As 4e² / n ---> 0 as n --> ∞, we deduce that f_n(x) converges uniformly to e^x on [0,2].