Series Convergent Criterion
\[\sum_{n} a_n \text{ converges} \iff s_{n} (n \in \mathbb{N}) \text{ converges } \iff s_n (n \in \mathbb{N}) \text{ is Cauchy}\] \[\forall \varepsilon > 0, \exists N \in \mathbb{N}, [k > N, l \ge 0 \implies \vert a_k + \dots + a_{k+l} \vert < \varepsilon ]\]since
\[\vert a_{k} + \dots + a_{k+l} \vert \le \vert a_k \vert + \dots + \vert a_{k+l} \vert\]thus if we have \(\sum_n \vert a_n \vert\) convergent, we are sure \(\sum_{n} a_n\) convergent. Since the partial sum of \(\sum_{n} \vert a_n \vert\) monotone, it converges if and only if its partial sum has a upper bound.
\[\exists M > 0 \forall n \in \mathbb{N} [\vert a_1 \vert + \dots \vert a_n \vert \le M]\]Notation: If \(a_n \ge 0 (n \in \mathbb{N})\), then we write \(\sum_{n} a_n < \infty\) to mean that \(\sum_n a_n\) converges. We write \(\sum_{n} a_n = \infty\) to mean that it diverges.
Examples (1) \(\sum_{n=1}^{\infty} \frac{1}{n}\)
\[S_1 = 1\] \[S_2 = 1 + \frac{1}{2}\] \[S_4 = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} > 1 + \frac{1}{2} + \frac{1}{2}\] \[\begin{align} \forall m \in \mathbb{N}, S_{2^m} &= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{2^{m-1}+1} + \dots + \frac{1}{2^{m-1}+2^{m-1}}\\ & > 1 + \frac{m}{2} \end{align}\](2) \(\sum_{n=1}^{\infty} \frac{1}{n^2}\)
\[\frac{1}{n^2} < \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}\] \[\begin{align} S_n &= 1 + \frac{1}{2^2} + \dots + \frac{1}{n^2} < 1 + (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2}-\frac{1}{3}) + \dots + (\frac{1}{n-1} - \frac{1}{n})\\ & < 1 + 1 - \frac{1}{n} < 2 \end{align}\](3) \(\sum_{n=1}^{\infty} \dfrac{\sin(n)}{n^k}\), for \(k \in \mathbb{N}, k \ge 2\)
\[\vert \dfrac{\sin(1)}{1^k} \vert + \dots + \vert \dfrac{\sin(n)}{n^k} \vert < 1 + \dots + \frac{1}{n^2} < 2\]thus \(\sum_{n=1}^{\infty} \dfrac{\sin(n)}{n^k}\) converges.
Ex2. For \(a > 1\) and \(k \in \mathbb{N}\), show that \(\sum_{n=1}^{\infty} \frac{n^k}{a^n} < \infty\)
It is obvious all terms are non-negative. Similar from the exercise in Calculus 003, for any \(n > 2k+ 4 > k+2\)
\[a^n = (1+b)^n \ge \binom{n}{k+2} b^{k+2} \ge \dfrac{b^{k+2}}{(k+2)!}(n-k-1)^{k+2} \ge \dfrac{b^{k+2}}{2^{k+2}(k+2)!} n^{k+2}\]thus for any \(n > 2k + 4\)
\[\dfrac{n^k}{a^n} \le \dfrac{2^{k+2}(k+2)!}{b^{k+2}} \dfrac{1}{n^2}\]thus the partial sum has upper bound.
Def: Given a sequence \(a_n (n \in \mathbb{N})\), we say that
(1) \(\sum_{n} a_n\) converges absolutely if \(\sum_{n} \vert a_n \vert < \infty\).
(2) \(\sum_{n} a_n\) converges conditionally if \(\sum_{n} \vert a_n \vert = \infty\) but \(\sum_{n=1}^{\infty} a_n\) converges.
Comparison Test
If \(a_n, b_n (n \in \mathbb{N}) \ge 0\)
(1) If we have
\[\exists C > 0 \text{ and } N \in \mathbb{N} [n \ge N \implies a_n \le C b_n]\]then
\[\sum_n b_n < \infty \implies \sum_n a_n < \infty\]Proof:
Since they are non-negative series, their convergence are equivalent to that the partial sum has upper bound. From the partial sum of \(b_n\) has upper bound, it is obvious that the partial sum of \(a_n\) has upper bound.
(2) If we know
\[\lim_{n\to\infty} \dfrac{a_n}{b_n} \text{ exists }\]then
\[\exists C > 0 \text{ and } N \in \mathbb{N} [n \ge N \implies a_n \le C b_n]\]Proof:
Let \(l = \lim_{n\to\infty} \dfrac{a_n}{b_n}\) and \(\varepsilon=1\)
\[\exists N \in \mathbb{N} [n \ge N \implies l-1 < \dfrac{a_n}{b_n} < l+1]\] \[\exists N \in \mathbb{N} [n \ge N \implies a_n < (l+1)b_n]\]where \(l+1\) must be positive \((C > 0)\), since \(l \ge 0\).
If \(\lim_{n\to\infty} \dfrac{a_n}{b_n}\) exists and not equal \(0\), it is easy to see \(\sum_{n} a_n < \infty \iff \sum_{n} b_n < \infty\), since \(a_n, b_n\) can be bounded by each other.
- Actually, \(\lim_{n\to\infty} \dfrac{a_n}{b_n}\) exists can be replaced by \(\overline{\lim_{n\to\infty}} \dfrac{a_n}{b_n}\) exists.
Proof: Let \(u_n = \sup \{\dfrac{a_m}{b_m} \vert m \ge n\}\), \(\lim_{n\to\infty}u_n = l\).
\[\exists N \in \mathbb{N} [u_N < l+1]\] \[\exists N \in \mathbb{N} [l+1 \text{ is an upper bound of } \{\dfrac{a_m}{b_m} \vert m \ge N\}]\] \[\exists N \in \mathbb{N} [n \ge N \implies \dfrac{a_n}{b_n} \le l + 1]\]Ratio and Root Tests
Exercise 3 (the ratio and the root tests):
Let \(a_n (n \in \mathbb{N}) \ge 0\). Then
(1) If \(\lim_{n\to\infty} \dfrac{a_{n+1}}{a_n} < 1\), then
\[\exists N \in \mathbb{N} \text{ and } C < 1 [n \ge N \implies a_{n+1} \le C a_n]\]then, \(\sum_{n} a_n < \infty\)
Proof: the idea is that
\[\exists N \in \mathbb{N} \text{ and }0 < C < 1 [n \ge N \implies a_{n} \le C^{n-N} a_N]\]Since the series \({a_1, a_2, \dots, a_{N-1}, C^{0} a_{N}, C^{1} a_{N}, \dots}\) converges, we know \(a_n (n\in\mathbb{N})\) converges.
Similarly, if \(\lim_{n\to\infty} \dfrac{a_{n+1}}{a_n} > 1\), we have \(\sum_n a_n = \infty\).
If \(\lim_{n\to\infty} \dfrac{a_{n+1}}{a_n} = 1\), it may converge or diverge.
(2) If \(\lim_{n\to\infty} (a_n)^{1/n} < 1\), then
\[\exists N \in \mathbb{N} \text{ and } C < 1 [n \ge N \implies a_n \le C^n]\]thus \(\sum_{n} a_n < \infty\).
If \(\lim_{n\to\infty} (a_n)^{1/n} > 1\), then \(\sum_{n} a_n = \infty\).
Alternating Series
Question: does \(\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n}\) converges (conditionally)?
Def: A series \(\sum_n a_n\) is an alternating series if
\[\exists b_n > 0 (n \in \mathbb{N}) \text{ s.t. } a_n = (-1)^{n-1} b_n (n \in \mathbb{N})\]Confusion: \(b_n > 0\) or \(b_n \ge 0\)?
Leibniz’s Criterion
If \(\sum_n a_n\) is an alternating series, and \(b_n (=\vert a_n \vert) \searrow 0\) as \(n \to \infty\) (monotonically non-increasing, and converges to \(0\)), then \(\sum_{n} a_n\) converges.
Proof:
We know \(b_n = (-1)^{n-1} a_n\). Consider the tail
\[\vert a_k + \dots + a_{k+l} \vert = \vert (-1)^{k-1} (b_k - b_{k+1} + \dots + (-1)^l b_{k+l}) \vert\]It is easy to see that \(b_k - b_{k+1} + \dots + (-1)^{l} b_{k+l} \ge 0\), no matter \(l\) is even or odd.
\[\begin{align} &\vert a_k + \dots + a_{k+l} \vert = \vert (-1)^{k-1} (b_k - b_{k+1} + \dots + (-1)^l b_{k+l}) \vert\\ =& b_k - b_{k+1} + \dots + (-1)^l b_{k+l} \le b_k \end{align}\]It is smaller than \(b_k\) since you can calculate it as \(b_k - (b_{k+1}-b_{k+2}) - \dots\), no matter \(l\) is even or odd.
Plus we know \(\lim_{n\to\infty} b_n = 0\), the tail can be controlled, and the series \(\sum_{n} a_n\) converges.