Let \(f_n \colon [0,1) \to \mathbb{R}\) be the function defined by \[f_n(x):= \sum_{k=1}^n \frac{x^k}{1+x^k}.\]
Prove that \(f_n\) converges to a function \(f \colon [0,1) \to \mathbb{R}\).
Prove that for every \(0 < a < 1\) the convergence is uniform on \([0,a]\).
Prove that \(f\) is differentiable on \((0,1)\).
Solution
1. Pointwise convergence on \([0,1)\).
for \(x \in [0,1)\), \(0 \le \frac{x^k}{1+x^k} \le x^k\), and \(\sum_k x^k\) is a convergent geometric series; hence \(\sum_k \frac{x^k}{1+x^k}\) converges, defining \(f(x)\).
2. Uniform convergence on \([0,a]\) for \(0 < a < 1\).
for \(x \le a\), \(\frac{x^k}{1+x^k} \le a^k\), and \(\sum_k a^k\) converges (geometric, \(a < 1\)); by the Weierstrass \(M\)-test the convergence is uniform on \([0,a]\).
3. \(f\) is differentiable on \((0,1)\).
each term \(g_k(x) = \frac{x^k}{1+x^k}\) is differentiable with \(g_k'(x) = \frac{kx^{k-1}}{(1+x^k)^2}\).
4. Q.E.D.