In this chapter we will learn the theory of "infinite sums".
Definition. Let \((a_m)_{m=0}^\infty\) be a sequence. Then the sequence \((s_m)_{m=0}^\infty\) defined by
\(s_m = \sum_{k=0}^m a_k, \quad m \ge 0\),
is called a series, and \(s_m\) is called its m-th partial sum. The series \((s_m)_{m=0}^\infty\) is also denoted by the formal expression
\(\sum_{k=0}^\infty a_k\).
If, however, the sequence \((s_m)_{m=0}^\infty\) has a limit, then we denote this limit by the same symbol
\(\sum_{k=0}^\infty a_k = \lim_{m \to \infty} \sum_{k=0}^m a_k\).
In this case we call the series convergent, and \(\sum_{k=0}^\infty a_k\) is called its sum.
If the series is not convergent, we call it divergent.
Remark Neither a series nor its partial sums have to start with index 0, e.g., we will regularly consider series of the form \(\sum_{m=1}^\infty a_m\).
The name of the index is not important either.
Examples. i) Let \(a_n := x^n, n \ge 0\), where \(x \in \mathbb{R}\) is a fixed number. Then
\(\sum_{k=0}^\infty x^k\) is called the geometric series.
If \(|x| < 1\), then we have
\(\lim_{m \to \infty} \sum_{k=0}^m x^k = \frac{1}{1-x}\), so the series is convergent with \(\sum_{k=0}^\infty x^k = \frac{1}{1-x}\).
If \(x \in \mathbb{R} \setminus (-1, 1)\), then the sequence of partial sums \(s_m = \sum_{k=0}^m x^k, m \ge 0\), is divergent.
ii) For the sequences \(a_m := \frac{1}{m!}, m \ge 0\), and \(b_m := \frac{1}{m^2}, m \ge 1\), we have already seen that
\(s_m = \sum_{k=0}^m a_k = \sum_{k=0}^m \frac{1}{k!}, m \ge 0\), and
\(t_m = \sum_{k=1}^m b_k = \sum_{k=1}^m \frac{1}{k^2}, m \ge 1\), are
bounded increasing sequences. Hence, by Thm 6, both series
\(\sum_{k=0}^\infty \frac{1}{k!}\) and \(\sum_{k=1}^\infty \frac{1}{k^2}\) are convergent.
Theorem 12. If the series \(\sum_{k=0}^\infty a_k\) is convergent, then
\(\lim_{m \to \infty} a_m = 0\).
Proof. Let us denote the sum of the series by s.
\(\forall m \ge 1: s_m - s_{m-1} = a_m\)
\(\implies \lim_{m \to \infty} a_m = \lim_{m \to \infty} (s_m - s_{m-1}) = s - s = 0\).
Theorem 13. Let \(\sum_{k=0}^\infty a_k\) and \(\sum_{k=0}^\infty b_k\) be convergent series and \(\alpha, \beta \in \mathbb{R}\). Then the series \(\sum_{k=0}^\infty (\alpha a_k + \beta b_k)\) is convergent with sum
\(\sum_{k=0}^\infty (\alpha a_k + \beta b_k) = \alpha \sum_{k=0}^\infty a_k + \beta \sum_{k=0}^\infty b_k\).
Proof. Using Thm 3 we obtain \(\lim_{m \to \infty} \sum_{k=0}^m (\alpha a_k + \beta b_k)\)
\(= \lim_{m \to \infty} \left\{ \alpha \sum_{k=0}^m a_k + \beta \sum_{k=0}^m b_k \right\}\)
\(= \alpha \lim_{m \to \infty} \sum_{k=0}^m a_k + \beta \lim_{m \to \infty} \sum_{k=0}^m b_k\)
\(= \alpha \sum_{k=0}^\infty a_k + \beta \sum_{k=0}^\infty b_k\).
Theorem 14 (Cauchy's Condensation Test)
Let \((a_m)_{m=1}^\infty\) be a decreasing sequence with \(a_m \ge 0\) for \(m \in \mathbb{N}\). Then the series \(\sum_{k=1}^\infty a_k\) is convergent if and only if the series \(\sum_{k=0}^\infty 2^k a_{2^k}\) is convergent.
Proof. "\(\implies\)" Let us denote the sum of the series \(\sum_{k=1}^\infty a_k\) by s. Then, for every \(m \in \mathbb{N}\), we have
\(s = \sum_{k=1}^\infty a_k \ge \sum_{k=1}^{2^m} a_k = a_1 + a_2 + (a_3 + a_4)\)
\(+ (a_5 + \cdots + a_8)\)
\(+ \cdots + (a_{2^{m-1}+1} + \cdots + a_{2^m})\)
\(\ge \frac{a_1}{2} + a_2 + 2 a_4 + 4 a_8 + \cdots + 2^{m-1} a_{2^m}\)
\(\implies a_1 + 2 a_2 + 4 a_4 + \cdots + 2^m a_{2^m} \le 2s\).
It follows by Thm 6, that \((\sum_{k=1}^m 2^k a_{2^k})_{m=1}^\infty\) is convergent.
"\(\impliedby\)" For \(m, n \in \mathbb{N}\) with \(n \le 2^m\) we have
\(\sum_{k=1}^n a_k \le \sum_{k=1}^{2^m} a_k \le a_1 + (a_2+a_3) + (a_4+\cdots+a_7)\)
\(+ \cdots + (a_{2^m} + \cdots + a_{2^{m+1}-1})\)
\(\le a_1 + 2 a_2 + 4 a_4 + \cdots + 2^m a_{2^m} \le \sum_{k=0}^\infty 2^k a_{2^k}\).
Hence, by Thm 6, \((\sum_{k=1}^m a_k)_{m=1}^\infty\) is convergent.
Example. For any \(\alpha \in \mathbb{R}\) we have that the series \(\sum_{m=1}^\infty \frac{1}{m^\alpha}\) is convergent if and only if \(\alpha > 1\):
\(\alpha \le 0\): If \(\alpha \le 0\), we have \(\frac{1}{m^\alpha} \not\to 0\), so the series cannot be convergent according to Thm 12. If \(\alpha > 0\), then according to Thm 14,
\(\sum_{m=1}^\infty \frac{1}{m^\alpha}\) is convergent \(\iff \sum_{m=0}^\infty 2^m \frac{1}{2^{m\alpha}}\) is convergent.
We have \(\sum_{m=0}^\infty 2^m \frac{1}{2^{m\alpha}} = \sum_{m=0}^\infty \left( \frac{1}{2^{\alpha-1}} \right)^m\), which
is convergent if and only if \(\frac{1}{2^{\alpha-1}} < 1 \iff \alpha> 1\).
In particular, the series \(\sum_{m=1}^\infty \frac{1}{m}\) is divergent. This series is called the harmonic series.
Moreover, the function \(\zeta: (1, \infty) \to \mathbb{R}\), defined by \(\zeta(x) = \sum_{m=1}^\infty \frac{1}{m^x}\), is called Riemann zeta function.
Theorem 15 (Dirichlet's Test)
Let \((a_m)_{m=0}^\infty\) be a sequence such that the sequence \(s_m = \sum_{k=0}^m a_k, m \ge 0\), is bounded, and let \((b_m)_{m=0}^\infty\) be a decreasing sequence converging to 0. Then the series \(\sum_{k=0}^\infty a_k b_k\) is convergent.
Proof. We know \(s_m\) is bounded
\(\implies \exists M > 0 \quad \forall m \ge 0: |\sum_{k=0}^m a_k| \le M\).
We use Abel's summation formula: For \(n > m \ge 0\)
\(\sum_{k=m+1}^n a_k b_k = b_{n+1} \sum_{k=m+1}^n a_k + \sum_{k=m+1}^n \left( \sum_{\nu=m+1}^k a_\nu \right) (b_k - b_{k+1}) \quad (*)\)
\(\sum_{k=m+1}^n (\sum_{\nu=m+1}^k a_\nu) (b_k - b_{k+1}) = \sum_{k=m+1}^n (\sum_{\nu=m+1}^k a_\nu) b_k - \sum_{k=m+1}^n (\sum_{\nu=m+1}^k a_\nu) b_{k+1}\)
\(= a_{m+1} b_{m+1} + \sum_{k=m+2}^n (\sum_{\nu=m+1}^k a_\nu) b_k - \sum_{k=m+2}^{n+1} (\sum_{\nu=m+1}^{k-1} a_\nu) b_k\)
\(= a_{m+1} b_{m+1} + \sum_{k=m+2}^n a_k b_k - (\sum_{\nu=m+1}^n a_\nu) b_{n+1}\)
\(= \sum_{k=m+1}^n a_k b_k - b_{n+1} \sum_{k=m+1}^n a_k\).
We now show that the sequence \(t_m = \sum_{k=0}^m a_k b_k, m \ge 0\), is a Cauchy sequence.
Let \(\epsilon > 0\). We know \(b_n \to 0\), so
\(\exists N \in \mathbb{N} \quad \forall n \ge N : b_n < \frac{\epsilon}{2M}\).
For \(n > m \ge N\) we have
\(|t_n - t_m| = |\sum_{k=m+1}^n a_k b_k|\)
\(\le b_{n+1} |\sum_{k=m+1}^n a_k| + \sum_{k=m+1}^n |\sum_{\nu=m+1}^k a_\nu| (b_k - b_{k+1})\) (by \((*)\))
\(= |s_n - s_m| = |s_k - s_m|\)
\(\le 2M (b_{n+1} + \sum_{k=m+1}^n (b_k - b_{k+1})) = 2M b_{m+1} < \epsilon\).
Theorem 16 (Leibniz's Test)
Let \((b_m)_{m=0}^\infty\) be a decreasing sequence converging to 0. Then the series \(\sum_{m=0}^\infty (-1)^m b_m\) is convergent.
Proof. Set \(a_n := (-1)^n, n \ge 0\), and use Thm 15.
Example. The alternating harmonic series
\(\sum_{m=1}^\infty \frac{(-1)^{m+1}}{m}\) is convergent by Thm 16.
Definition. Let \((a_m)_{m=0}^\infty\) be a sequence. We say that the series \(\sum_{m=0}^\infty a_m\) is absolutely convergent if \(\sum_{m=0}^\infty |a_m|\) is convergent.
Remarks. i) The series \(\sum_{m=1}^\infty \frac{(-1)^{m+1}}{m}\) is convergent but not absolutely convergent.
ii) In the case \(a_n \ge 0, n \ge 0\), the notions of convergence and absolute convergence coincide.
Theorem 17. Every absolutely convergent series is convergent.
Proof. Let \(\sum_{m=0}^\infty a_m\) be absolutely convergent. We want to show that \(s_m = \sum_{k=0}^m a_k, m \ge 0\), is convergent. Let \(\epsilon > 0\). We know by Thm 8 that \(\sum_{k=0}^\infty |a_k|, m \ge 0\), is a Cauchy sequence, hence
\(\exists N \in \mathbb{N} \quad \forall n \ge m \ge N : \sum_{k=m+1}^n |a_k| < \epsilon\).
\(\implies \forall n \ge m \ge N : |s_n - s_m| = |\sum_{k=m+1}^n a_k| \le \sum_{k=m+1}^n |a_k| < \epsilon\).
\(\implies (s_m)_{m=0}^\infty\) is Cauchy \(\implies\) (Thm 8) \((s_m)\) convergent.
Theorem 18. If \(\sum_{m=0}^\infty a_m\) and \(\sum_{m=0}^\infty b_m\) are absolutely convergent and \(\alpha, \beta \in \mathbb{R}\), then the series \(\sum_{m=0}^\infty (\alpha a_n + \beta b_n)\) is absolutely convergent.
Proof. Follows immediately using the triangle inequality.
Theorem 19 (Comparison Test)
Let \((a_m)_{m=0}^\infty\) and \((b_m)_{m=0}^\infty\) be sequences and \(N \in \mathbb{N}\).
i) If \(\sum_{m=0}^\infty b_m\) is absolutely convergent and if there exists a constant \(K > 0\) such that \(|a_m| \le K |b_m|\) for all \(m \ge N\), then \(\sum_{m=0}^\infty a_m\) is absolutely convergent.
ii) If \(\sum_{m=0}^\infty |a_m|\) is divergent and if there is a constant \(K > 0\) such that \(|b_m| \ge K |a_m|\) for all \(m \ge N\), then \(\sum_{m=0}^\infty |b_m|\) is divergent.
iii) If \(a_n, b_n > 0, n \ge 0\), and \(\lim_{n \to \infty} \frac{a_n}{b_n} = c\) with \(c \in (0, \infty)\), then
\(\sum_{m=0}^\infty a_n\) is convergent \(\iff \sum_{m=0}^\infty b_n\) is convergent.
Proof. i) For \(m \ge N\) we have
\(\sum_{k=0}^m |a_k| = \sum_{k=0}^{N-1} |a_k| + \sum_{k=N}^m |a_k| \le \sum_{k=0}^{N-1} |a_k| + K \sum_{k=N}^m |b_k|\)
\(\le \sum_{k=0}^{N-1} |a_k| + K \sum_{k=N}^\infty |b_k|\),
so the sequence \(\sum_{k=0}^m |a_k|, m \ge 0\), is increasing and bounded
\(\implies\) (Thm 6) \(\sum_{m=0}^\infty |a_m|\) is convergent.
ii) For \(m \ge N\) we have
\(\sum_{k=0}^m |b_k| = \sum_{k=0}^{N-1} |b_k| + \sum_{k=N}^m |b_k| \ge \sum_{k=0}^{N-1} |b_k| + K \sum_{k=N}^m |a_k|\)
\(\to \infty, m \to \infty\).
iii) We know \(\lim_{n \to \infty} \frac{a_n}{b_n} = c \in (0, \infty)\)
\(\implies \exists N_1 \in \mathbb{N} \quad \forall n \ge N_1 : \frac{1}{2}c \le \frac{a_n}{b_n} \le \frac{3}{2}c\)
\(\implies \forall n \ge N_1 : \frac{1}{2}c b_n \le a_n \le \frac{3}{2}c b_n\).
Now the statement follows from i) and ii).
Examples. i) The series \(\sum_{m=1}^\infty \frac{\sqrt{m}}{m^2+1}\) is convergent by part i) of Thm 19 :
\(\forall m \ge 1: \frac{\sqrt{m}}{m^2+1} \le \frac{\sqrt{m}}{m^2} = \frac{1}{m^{3/2}}\),
and \(\sum_{m=1}^\infty \frac{1}{m^{3/2}}\) is convergent.