and \(\sum_{m=1}^\infty \frac{1}{m^{3/2}}\) is convergent.
ii) The series \(\sum_{m=1}^\infty \frac{m}{m^2+1}\) is divergent by part ii) of Thm 19:
\(\forall m \ge 1: \frac{m}{m^2+1} \ge \frac{m}{m^2+m^2} = \frac{1}{2m}\),
and \(\sum_{m=1}^\infty \frac{1}{m}\) is divergent.
Theorem 20. (Root Test)
Let \((a_m)_{m=0}^\infty\) be a sequence.
i) If \(\limsup_{m \to \infty} \sqrt[m]{|a_m|} < 1\), then the series \(\sum_{m=0}^\infty a_m\) is absolutely convergent.
ii) If \(\limsup_{m \to \infty} \sqrt[m]{|a_m|} > 1\), then the series \(\sum_{m=0}^\infty a_m\) is divergent.
Proof. i) We know \(\limsup_{m \to \infty} \sqrt[m]{|a_m|} < 1\)
\(\implies \exists q \in [0, 1), N \in \mathbb{N} \quad \forall m \ge N: \sqrt[m]{|a_m|} \le q\)
\(\implies \forall m \ge N: |a_m| \le q^m\), and \(\sum_{m=0}^\infty q^m\) is convergent.
Hence, by Thm 19, \(\sum_{m=0}^\infty a_m\) is abs. convergent.
ii) We know \(\limsup_{m \to \infty} \sqrt[m]{|a_m|} > 1\)
\(\implies \sqrt[m]{|a_m|} \ge 1\) for infinitely many m
\(\implies |a_m| \ge 1\) for infinitely many m
\(\implies a_m \not\to 0 \implies \sum_{m=0}^\infty a_m\) is divergent by Thm 12.
Example. The series \(\sum_{m=1}^\infty (1+\frac{1}{m})^{-m^2}\) is abs. convergent:
\(\limsup_{m \to \infty} ((1+\frac{1}{m})^{-m^2})^{1/m} = \limsup_{m \to \infty} (1+\frac{1}{m})^{-m}\)
\(= \limsup_{m \to \infty} \frac{1}{(1+\frac{1}{m})^m} = \lim_{m \to \infty} \frac{1}{(1+\frac{1}{m})^m} = \frac{1}{e} < 1\).
Theorem 21. (Ratio Test)
Let \((a_m)_{m=0}^\infty\) be a sequence and \(K \in \mathbb{N}\) such that \(a_m \neq 0\) for \(m \ge K\).
i) If \(\limsup_{m \to \infty} | \frac{a_{m+1}}{a_m} | < 1\), then the series \(\sum_{m=0}^\infty a_m\) is absolutely convergent.
ii) If \(\liminf_{m \to \infty} | \frac{a_{m+1}}{a_m} | > 1\), then the series \(\sum_{m=0}^\infty a_m\) is divergent.
Proof. i) We know \(\limsup_{m \to \infty} | \frac{a_{m+1}}{a_m} | < 1\)
\(\implies \exists q \in [0, 1), N \in \mathbb{N} \quad \forall m \ge N: | \frac{a_{m+1}}{a_m} | \le q\)
\(\implies \forall m \ge N \quad |a_{m+1}| \le q |a_m| \le q^2 |a_{m-1}| \le \dots \le q^{m+1-N} |a_N|\)
\(\implies \forall m \ge N+1: |a_m| \le q^m \frac{|a_N|}{q^N}\), and \(\sum_{k=0}^\infty q^m\) is convergent
\(\implies (\text{Thm 19}) \sum_{m=0}^\infty a_m\) is absolutely convergent.
ii) We know \(\liminf_{m \to \infty} | \frac{a_{m+1}}{a_m} | > 1\)
\(\implies \exists N \in \mathbb{N} \quad \forall m \ge N: | \frac{a_{m+1}}{a_m} | \ge 1\)
\(\implies \forall m \ge N: |a_{m+1}| \ge |a_m|\)
Together with \(a_N \neq 0\), it follows that \(a_m \not\to 0\), so \(\sum_{m=0}^\infty a_m\) is divergent by Thm 12.
Example. The series \(\sum_{m=1}^\infty \frac{m!}{m^m}\) is abs. convergent:
We set \(a_m := \frac{m!}{m^m}, m \ge 1\), then
\(| \frac{a_{m+1}}{a_m} | = \frac{(m+1)! m^m}{(m+1)^{m+1} m!} = (\frac{m}{m+1})^m = \frac{1}{(1+\frac{1}{m})^m}\)
\(\implies \limsup_{m \to \infty} | \frac{a_{m+1}}{a_m} | = \lim_{m \to \infty} \frac{1}{(1+\frac{1}{m})^m} = \frac{1}{e} < 1\).
Remark. As we can see from the series \(\sum_{m=1}^\infty \frac{1}{m}\) and \(\sum_{m=1}^\infty \frac{1}{m^2}\), the root test and the ratio test are inconclusive if \(\limsup_{m \to \infty} \sqrt[m]{|a_m|} = 1\) and \(\limsup_{m \to \infty} | \frac{a_{m+1}}{a_m} | = 1\), respectively.
Next we turn to multiplication of series.
Let us first consider finite sums \(\sum_{j=0}^M a_j\) and \(\sum_{k=0}^M b_k\). Multiplying them gives
\(S := (\sum_{j=0}^M a_j) (\sum_{k=0}^M b_k) = \sum_{j, k=0}^M a_j b_k\).
We can imagine S as the result of adding all entries of the matrix
\[ \begin{pmatrix} a_0 b_0 & a_0 b_1 & a_0 b_2 & \dots & a_0 b_M \\ a_1 b_0 & a_1 b_1 & a_1 b_2 & & \vdots \\ a_2 b_0 & a_2 b_1 & a_2 b_2 & & \\ \vdots & & & \ddots & \\ a_M b_0 & \dots & & & a_M b_M \end{pmatrix} \]
As the order of summation does not matter we can sum it
row-wise: \(S = \sum_{j=0}^M \sum_{k=0}^M a_j b_k\),
column-wise: \(S = \sum_{k=0}^M \sum_{j=0}^M a_j b_k\),
or along diagonals:
\(S = \sum_{k=0}^{2M} \sum_{j=0}^k a_j b_{k-j}\), where \(a_j := b_k := 0\) if \(j, k > M\).
Definition. For two series \(\sum_{k=0}^\infty a_k\) and \(\sum_{k=0}^\infty b_k\) the series \(\sum_{k=0}^\infty c_k\), with \(c_k = \sum_{j=0}^k a_j b_{k-j}\), is called their Cauchy product.
Theorem 22
If \(\sum_{k=0}^\infty a_k\) and \(\sum_{k=0}^\infty b_k\) are absolutely convergent, then their Cauchy product is absolutely convergent and
\(\sum_{k=0}^\infty (\sum_{j=0}^k a_j b_{k-j}) = (\sum_{j=0}^\infty a_j) (\sum_{k=0}^\infty b_k)\).
Proof. First we show the absolute convergence of the Cauchy product. To this end it is sufficient to show the boundedness
of the partial sums. For \(M \ge 0\) we have
\(\sum_{k=0}^M |c_k| \le \sum_{k=0}^M \sum_{j=0}^k |a_j| |b_{k-j}|\)
\(\le \sum_{j=0}^M |a_j| \sum_{k=0}^M |b_k|\)
\(= (\sum_{j=0}^M |a_j|) (\sum_{k=0}^M |b_k|) \le (\sum_{j=0}^\infty |a_j|) (\sum_{k=0}^\infty |b_k|)\).
In order to show that \(\sum_{k=0}^\infty c_k = (\sum_{j=0}^\infty a_j) (\sum_{k=0}^\infty b_k)\), we show
\(\lim_{M \to \infty} (\sum_{k=0}^{2M} c_k - (\sum_{j=0}^M a_j) (\sum_{k=0}^M b_k)) = 0\).
We have for \(M \ge 0\)
\(| \sum_{k=0}^{2M} (\sum_{j=0}^k a_j b_{k-j}) - \sum_{j,k=0}^M a_j b_k |\)
\(= | \sum_{j=0}^{M-1} \sum_{k=M+1}^{2M-j} a_j b_k + \sum_{j=M+1}^{2M} \sum_{k=0}^{2M-j} a_j b_k |\)
\(\le \sum_{j=0}^{M-1} |a_j| \sum_{k=M+1}^{2M-j} |b_k| + \sum_{j=M+1}^{2M} |a_j| \sum_{k=0}^{2M-j} |b_k|\)
\(\le \sum_{j=0}^\infty |a_j| \cdot \underbrace{\sum_{k=M+1}^\infty |b_k|}_{\to 0} + \underbrace{\sum_{j=M+1}^\infty |a_j|}_{\to 0} \cdot \sum_{k=0}^\infty |b_k| \to 0\), \(M \to \infty\).
Here we used \(\sum_{k=M+1}^\infty |b_k| = \sum_{k=0}^\infty |b_k| - \sum_{k=0}^M |b_k| \to 0\).
We now turn to "rearrangements".
Definition. If \((a_n)_{n=0}^\infty\) is a sequence and \(\varphi: \mathbb{N}_0 \to \mathbb{N}_0\) is a bijective mapping, then the sequence \((b_k)_{k=0}^\infty\) defined by \(b_k = a_{\varphi(k)}\), for \(k \ge 0\), is called a rearrangement of \((a_n)\).
It is not difficult to show that, if \(\lim_{n \to \infty} a_n = a\), then we also have \(\lim_{k \to \infty} a_{\varphi(k)} = a\).
For series, however, we must be very careful with rearrangements.
The famous Riemann rearrangement theorem tells us that, if \(\sum_{n=0}^\infty a_n\) is convergent but not absolutely convergent, then for every number \(s \in \mathbb{R}\) we can find a rearrangement \((a_{\varphi(k)})_{k=0}^\infty\) such that
\(\sum_{k=0}^\infty a_{\varphi(k)} = s\).
This, however, cannot happen with absolutely convergent series.
Theorem 23. Let \(\sum_{n=0}^\infty a_n\) be absolutely convergent. Then, for every rearrangement \((a_{\varphi(k)})_{k=0}^\infty\) the series \(\sum_{k=0}^\infty a_{\varphi(k)}\) is absolutely convergent and
\(\sum_{k=0}^\infty a_{\varphi(k)} = \sum_{n=0}^\infty a_n\).
Proof. Let \(\epsilon > 0 \implies \exists N \in \mathbb{N}: \sum_{k=N+1}^\infty |a_k| < \frac{\epsilon}{2}\).
Moreover, \(\varphi\) is bijective \(\implies \exists K \in \mathbb{N}: \{0, 1, \dots, N\} \subset \{ \varphi(0), \varphi(1), \dots, \varphi(K) \}\)
\(\implies \forall k > K: \varphi(k) > N\)
\(\implies \forall m \ge n > K : | \sum_{k=0}^m |a_{\varphi(k)}| - \sum_{k=0}^n |a_{\varphi(k)}| | = \sum_{k=n+1}^m |a_{\varphi(k)}| \le \sum_{k=N+1}^\infty |a_k| < \frac{\epsilon}{2} < \epsilon\).
\(\implies \sum_{k=0}^\infty a_{\varphi(k)}\) is absolutely convergent.
Now let us consider \(M \ge K\) (\(\ge N\)), then
\(| \sum_{k=0}^M a_{\varphi(k)} - \sum_{k=0}^M a_k | = | \sum_{k=0}^M a_{\varphi(k)} - \sum_{k=0}^N a_k - \sum_{k=N+1}^M a_k |\)
\(\le | \sum_{k=0, \varphi(k)>N}^M a_{\varphi(k)} | + \sum_{k=N+1}^\infty |a_k| \le 2 \sum_{k=N+1}^\infty |a_k| < \epsilon\).
We now turn to the most important class of series.
Definition. Let \((a_n)_{n=0}^\infty\) be a sequence and \(x_0 \in \mathbb{R}\). A power series is a series of the form
\(\sum_{n=0}^\infty a_n (x-x_0)^n, \quad x \in \mathbb{R}\),
where the \(a_n\)'s are called its coefficients and \(x_0\) its center.
Remarks. i) In many situations we have \(x_0 = 0\), and the power series takes the simpler form
\(\sum_{n=0}^\infty a_n x^n\).
ii) The partial sums of a power series
\(s_m(x) = \sum_{k=0}^m a_k (x-x_0)^k, \quad m \ge 0\),
are polynomials in x of degree \(\le m\).
iii) If \(a_n = 0\) for all \(n > N\) (\(N \in \mathbb{N}\) fixed), then the power series is convergent for every \(x \in \mathbb{R}\) with
\(\sum_{n=0}^\infty a_n (x-x_0)^n = \sum_{n=0}^N a_n (x-x_0)^n\),
which is a polynomial in x.