Taylor's Theorem

Weierstrass M-test. If you can find a sequence of non-negative numbers, $M_k$, such that $|u_k(z)|\le M_{k}, \forall z \in \gamma*$ and the series $\sum_{k=0}^\infin M_k$ converges, then the original series of functions $\sum_{k=0}^\infin u_k(z)$ converges uniformly on that domain.

Lemma 1. Let $\gamma$ be a path, let $U, u_0, u_1, \cdots u_n, \cdots$ be continuous on the trace of $\gamma, \gamma*$, and assume that $\sum_{k=0}^\infin u_k(z)$ converges to U(z) for all $z \in \gamma*$. Assume that there exist constants $M_k, k \in \mathbb{Z}, k \ge 0$ such that $\sum_{k=0}^\infin M_k$ converges and $|u_k(z)| \le M_k, \forall z \in \gamma*$. Then, $\sum_{k=0}^\infin \int_\gamma u_k(z)dz = \int_\gamma \sum_{k=0}^\infin u_k(z)dz = \int_\gamma U(z)dz$.

In calculus, we know that for a finite sum, the integral of the sum is the sum of the integrals (linearity of the integral): $\int_\gamma (u_1(z) + u_2(z)) dz = \int_\gamma u_1(z) dz + \int_\gamma u_2(z) dz$. However, this is not automatically true for an infinite sum. This lemma states that we can swap them, provided the series converges “nicely” enough (Weierstrass M-test).

Proof.

The strategy of the proof is to show that the error (the difference) between the full integral $\int U(z)dz$ and the $N$-th partial sum of the integrals $\sum_{k=0}^N \int u_k(z)dz$ must go to zero as $N$ gets large.

Let $U(z) = \sum_{k=0}^\infty u_k(z)$ be the final sum of the series, and let $U_N(z) = \sum_{k=0}^N u_k(z)$ be the $N$-th partial sum, $N = 0, 1, \cdots$. Our goal is to prove that $\int_\gamma U(z) dz = \sum_{k=0}^\infty \int_\gamma u_k(z) dz =[\text{Definition of a series}] \lim_{N \to \infty} \sum_{k=0}^N \int_\gamma u_k(z) dz$.

We want to show that the integral of the full sum is the limit of the sum of the integrals. Let’s look at the difference or “error” between the integral of the full sum and the N-th partial sum of the integrals:

$$ \begin{aligned} Error_N &=|\int_\gamma U(z)dz - \sum_{k=0}^N \int_\gamma u_k(z)dz| \\[2pt] &\text{Because } U_N(z) \text{ is a finite sum, we can use the linearity property to pull the sum outside the integral on the right side:}\\[2pt] &=|\int_\gamma U(z)dz - \int_\gamma \sum_{k=0}^N \left(u_k(z) \right)dz| \\[2pt] &=|\int_\gamma U(z)dz - \int_\gamma U_N(z)dz| \\[2pt] &\text{Again, by linearity, we can combine the two integrals:} \\[2pt] &=|\int_\gamma (U(z) - U_N(z))dz| \\[2pt] &\text{Apply the ML-Inequality, } |\int_\gamma f(z)dz| \le M \cdot length(\gamma) \text{, where lenght(γ) is the path length of the contour and M is the maximum value of |f| on it.} \\[2pt] &\lt sup_{z \in \gamma*}\{ |U(z) - U_N(z)| \} \cdot length(\gamma) \\[2pt] &U(z) - U_N(z) \text{ is the tail or remainder of the infinite series after the N-th term.} \\[2pt] &=sup_{z \in \gamma*}\{ |\sum_{k=N+1}^\infin u_k(z) \}| \cdot length(\gamma) \\[2pt] &\text{Triangle Inequality for infinite series.} \\[2pt] &=sup_{z \in \gamma*}\{ \sum_{k=N+1}^\infin |u_k(z)| \} \cdot length(\gamma) \\[2pt] &\le \sum_{k=N+1}^\infin M_k \cdot length(\gamma). \end{aligned} $$

We have found a bound for $|U(z) - U_N(z)|$ that is independent of z, $0 \le \text{Error}_N \le \text{length}(\gamma) \cdot \left( \sum_{k=N+1}^\infty M_k \right)$

We are given that $\sum_{k=0}^\infty M_k$ is a convergent series. A fundamental property of convergent series is that their tail must go to zero, $\lim_{N \to \infty} \left( \sum_{k=N+1}^\infty M_k \right) = 0$. As $N \to \infty$, both the right and left sides of our inequality goes to 0.

By the Squeeze Theorem, the error term in the middle must also go to zero: $\lim_{N \to \infty} \text{Error}_N = 0, \lim_{N \to \infty} \left| \int_\gamma U(z) dz - \sum_{k=0}^N \int_\gamma u_k(z) dz \right| = 0$

This is the formal definition of the limit, which means: $\int_\gamma U(z) dz = \lim_{N \to \infty} \sum_{k=0}^N \int_\gamma u_k(z) dz =[\text{Definition of an infinite series}] \sum_{k=0}^\infty \int_\gamma u_k(z) d$