Dominated Convergence Theorem for Series

In this post, we will cover a classical but nevertheless powerful result about interchanging the order of limit and sum.

Let $G$ be an abelian group. We will adopt additive notation for the group operation. Assume that $G$ is endowed with a norm $\|\cdot\|$ with respect to which $G$ is complete.

Theorem. Let $(a^{(n)}_k)_{n,k=1}^{\infty} \subseteq G$ and $(M_k)_{k=1}^{\infty} \subseteq [0, \infty)$ satisfy the following conditions:

1. $\| a^{(n)}_k \| \leq M_k$ holds for each $n$ and $k$;
2. $(a^{(n)}_k)_{n=1}^{\infty}$ converges in $G$ as $k \to \infty$ for each $n$;
3. $\sum_{k=1}^{\infty} M_k$ converges.

Then $\sum_{k=1}^{\infty} a^{(n)}_k$ converges for each $n$, and $$\lim_{n\to\infty} \sum_{k=1}^{\infty} a^{(n)}_k = \sum_{k=1}^{\infty} \lim_{n\to\infty} a^{(n)}_k.$$

Remark. The completeness of $G$ is not necessary provided all the sums appearing in the equality converge in $G$.

Proof. For each $n$, let $S^{(n)} := \sum_{k=1}^{\infty} a^{(n)}_k$. Then for $p \lt q$, we have $$\left\| \sum_{k=p+1}^{q} a^{(n)}_k \right\| \leq \sum_{k=p+q}^{q} \| a^{(n)}_k\| \leq \sum_{k=p+1}^{q} M_k.$$ Since this bound converges to $0$ as $p, q \to \infty$, the partial sums of $S^{(n)}$ define a Cauchy sequence in $G$ for each $n$. Then the completeness of $G$ ensures that the series $S^{(n)}$ converges in $G$ for each $n$.

Define the sequence $(a_k)_{k=1}^{\infty}$ in $G$ by $a_k := \lim_{n\to\infty} a^{(n)}_k$ for each $k$. Then it is clear that $\| a_k \| \leq M_k$, and so, the sum $S := \sum_{k=1}^{\infty} a_k$ converges by the same reasoning. Finally, for each fixed $p$, $$\begin{align*} \| S^{(n)} - S \| &\leq \left\| S^{(n)} - \sum_{k=1}^{p} a_k \right\| + \left\| \sum_{k=1}^{p} a_k - S \right\| \\ &\leq \sum_{k=1}^{p} \| a^{(n)}_k - a_k \| + \sum_{k=p+1}^{\infty} \| a^{(n)}_k \| + \sum_{k=p+1}^{\infty} \| a_k \| \\ &\leq \sum_{k=1}^{p} \| a^{(n)}_k - a_k \| + 2 \sum_{k=p+1}^{\infty} M_k. \end{align*}$$ Letting $\limsup$ as $n \to \infty$, we obtain the bound $$\limsup_{n\to\infty} \| S^{(n)} - S \| \leq 2 \sum_{k=p+1}^{\infty} M_k.$$ Since the left-hand side does not depend on $p$, letting $p \to \infty$ shows that the limsup is in fact $0$, proving $S^{(n)} \to S$ in $G$ as $n \to \infty$. This conclusion is equivalent to the desired assertion, hence the proof is done. $\square$

Example. We show that $$\sum_{k=1}^{\infty} \frac{2^k}{k} = 0$$ in $\mathbb{Q}_2$. Indeed, for $n \geq 1$ we have $$\begin{align*} 0 = \frac{1 - (1 - 2)^{2^n}}{2^n} &= \frac{1}{2^n} \sum_{k=1}^{\infty} (-1)^{k-1} \binom{2^n}{k} 2^k \\ &= \sum_{k=1}^{\infty} (-1)^{k-1} \binom{2^n - 1}{k - 1} \frac{2^k}{k} \\ &= \sum_{k=1}^{\infty} \left[ \prod_{l=1}^{k-1} \left( 1 - \frac{2^n}{l} \right) \right] \frac{2^k}{k}. \end{align*}$$ It is easy to check that the $|\cdot|_2$-norm of the $k$th term is uniformly bounded by $2^{-k}$ for all $n$ and $k$, and $\prod_{l=1}^{k-1} \left( 1 - \frac{2^n}{l} \right) \to 1$ in $\mathbb{Q}_2$ as $n \to \infty$ for each $k$. So by DCT, it follows that $$0 = \sum_{k=1}^{\infty} \left[ \lim_{n \to \infty} \prod_{l=1}^{k-1} \left( 1 - \frac{2^n}{l} \right) \right] \frac{2^k}{k} = \sum_{k=1}^{\infty} \frac{2^k}{k}.$$

Remark. The above result can be interpreted in terms of logarithm as follows: Note that the logarithm $\log(z) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} z^n$ converges in $\mathbb{Q}_2$ whenever $|z|_2 < 1$, and moreover, $$\log(1+z) + \log(1+w) = \log((1+z)(1+w)),$$ which initially holds as an identity between formal power series, extends to all $z, w \in \mathbb{Q}_2$ with $|z|_2 < 1$ and $|w|_2 < 1$. So, $$\sum_{n=1}^{\infty} \frac{2^n}{n} = - \log(1-2) = - \log (-1) = - \frac{1}{2} \log((-1)^2) = -\frac{1}{2}\log 1 = 0.$$