Homework 12

Fix \(\varepsilon>0\) and for each \(n\in\mathbb{N}\) we denote \[ E_n = \left\{x\in[a,b]: f_n(x) > f(x) - \varepsilon\right\}\] and observe that each \(E_n\), being the preimage of the continuous function \(f - f_n\) on the open interval\((-\varepsilon,\varepsilon)\), is open. By pointwise convergence \(\left\{E_n\right\}_{n\in\mathbb{N}}\) forms an open cover of \([a,b]\), which is compact, so let \(E_{n_1}, E_{n_2}, \dots, E_{n_k}\) be a finite subcover. Then let \(N = \max(n_1, n_2, \dots, n_k)\) and whenever \(n>N\), it follows from covering and \(f_n \leq f_{n+1}\) pointwise that \(f(x) - f_n(x) < \varepsilon\).

Homework 13

Q1

Part (i)

If RHS finite then we just use the standard Abel theorem. If RHS infinite as each \(a_n \geq 0\) it follows that \[ \sum_{n=1}^\infty a_nR = +\infty. \] First without loss of generality assume \(R = 1\), the proof for the infinite case resembles that of the finite case. Let \(S_n = \sum_{k=0}^n a_k\) and we assume that \(S_n \to \infty\). For any \(M>0\) there exists \(N\) such that \[n\geq N\implies S_n > M.\] Summing by parts, whenever \(n>N\) we have \[\begin{align*} \sum_{k=0}^n a_kx^k &= S_nx^n + \sum_{k=0}^{n-1}S_k(x^k - x^{k+1}) \\ &= S_nx^n + (1-x) \sum_{k=0}^{n-1}S_k x^k \\ &> M x^n + (1-x)\left[ \sum_{k=0}^{N-1}S_k x^k + \sum_{k=N}^n S_k x^k \right] \\ &> M x^n + (1-x)\left[ \sum_{k=0}^{N-1}S_k x^k + M \sum_{k=N}^n x^k \right] \\ &= M x^n + (1-x)\left[ \sum_{k=0}^{N-1}S_k x^k + M \frac{x^N-x^n}{1-x} \right] \\ &= M x^N + (1-x)\sum_{k=0}^{N-1} S_k x^k \end{align*}\] and it follows that \[ \sum_{n=0}^\infty a_nx^n \geq M x^N + (1-x)\sum_{k=0}^{N-1}S_k x^k. \] Hence \(\lim_{x\to1^-} \sum_{n=0}^\infty a_nx^n \geq M\), and since \(M\) was arbitrary the limit is actually \(+\infty\) as desired.

Part (ii)

To show \[\sum_{n=1}^\infty \frac1n = +\infty\] by part (i) we just need to show that \[\lim_{x\to1^-} \sum_{n=1}^\infty \frac{x^n}{n} = +\infty\] as this power series has radius of convergence \(R=1\). Denote the power series as \[G(x) = \sum_{n=1}^\infty \frac{x^n}{n}\] and we have \[G'(x) = \sum_{n=0}^\infty x^n = \frac{1}{1-x}\] for all \(x\in(-R,R)\). Therefore \[\lim_{x\to 1^-} G(x) = \lim_{x\to 1^-} \int_0^x \frac1{1-t}\ dt = \lim_{x\to 1^-} -\ln(1-x) = +\infty. \]

Q2

Let \(\varepsilon>0\), as the limit \[\lim_{x\to 1^-} \sum_{n=1}^\infty a_nx^n = A\] we can find \(0<c<1\) near enough to \(1\) such that \[\sum_{n=1}^\infty a_nc^n > A-\frac\varepsilon 2 \] then as the finite sum converges we can find \(N\) large enough such that \[\sum_{n=N+1}^\infty a_nc^n < \frac\varepsilon 2 \] and as a result for any \(M>N\), \[\sum_{n=1}^M a_nc^n > A - \varepsilon\] Now we have \[ A = \lim \sum a_nx^n \geq \lim \sum_{n=1}^M a_nx^n = \sum_{n=1}^M a_n \geq \sum_{n=1}^M a_nc^n > A-\varepsilon\] which means \(\sum_{n=1}^\infty a_n = A\) too.

The condition on the coefficients was used a lot and cannot be removed. For the counter example consider \(a_n = (-1)^n\), then the series \[\sum_{n=0}^\infty (-x)^n\] has radius of convergence \(R=1\). However for all \(-1 < x < 1\) this series converges to \(\frac{1}{1+x}\) and the limit as \(x\to1\) from the left is \(\frac12\), but the sum \(\sum_{n=0}^\infty (-1)^n\) does not converge.

Q3

As \(a_n\) is strongly dominated by \(\frac1{n^{1+\alpha}}\) where \(\alpha>0\), then \[\lim_{n\to\infty} n^\alpha n a_n = 0\] from this it follows that \(\lim_{n\to\infty} n a_n = 0\) too and we can directly apply Tauber’s theorem to get the result.