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Section 3.3 More Convergence Tests for Series

The tests in Section 3.2 are enough for many examples, but two further tools are often useful. The first concerns products of convergent series, and the second sharpens the ratio test in the borderline case when the quotient of consecutive terms tends to \(1\text{.}\)

Subsection 3.3.1 Cauchy Products

Given two series
\begin{equation*} \sum_{n=0}^\infty a_n \qquad \text{and} \qquad \sum_{n=0}^\infty b_n, \end{equation*}
their Cauchy product is the series
\begin{equation*} \sum_{n=0}^\infty c_n, \qquad c_n:=\sum_{k=0}^n a_kb_{n-k}. \end{equation*}
The coefficient \(c_n\) is the sum of all products whose indices add to \(n\text{.}\)

Proof.

Example 3.3.2.

Let \(|x|<1\text{.}\) The Cauchy product of
\begin{equation*} \sum_{n=0}^\infty x^n \qquad \text{with itself} \end{equation*}
has coefficients
\begin{equation*} c_n=\sum_{k=0}^n x^k x^{n-k}=(n+1)x^n. \end{equation*}
Since the geometric series converges absolutely for \(|x|<1\) by Proposition 3.2.1, Mertens’ theorem gives
\begin{equation*} \sum_{n=0}^\infty (n+1)x^n = \left(\sum_{n=0}^\infty x^n\right)^2 = \frac{1}{(1-x)^2}. \end{equation*}

Example 3.3.3.

The Cauchy product of two conditionally convergent series may diverge. Consider
\begin{equation*} a_n=b_n=\frac{(-1)^{n+1}}{\sqrt{n}} \qquad (n \ge 1). \end{equation*}
By the alternating series test (Proposition 3.2.15), the series \(\sum_{n=1}^\infty a_n\) converges. It is not absolutely convergent, because \(\sum 1/\sqrt{n}\) diverges by the \(p\)-series test (Proposition 3.2.5).
If \(c_n=\sum_{i+j=n}a_ib_j\) for \(n \ge 2\text{,}\) then
\begin{equation*} |c_n| = \left|\sum_{i+j=n}\frac{(-1)^{i+j}}{\sqrt{ij}}\right| = \sum_{i+j=n}\frac{1}{\sqrt{ij}}. \end{equation*}
For \(i,j \ge 1\) with \(i+j=n\text{,}\) we have \(ij \lt (i+j)^2=n^2\text{,}\) so
\begin{equation*} \frac{1}{\sqrt{ij}} > \frac{1}{n}. \end{equation*}
Since there are \(n-1\) such pairs, it follows that
\begin{equation*} |c_n| > \frac{n-1}{n} \to 1. \end{equation*}
Thus \(c_n \not\to 0\text{,}\) and the Cauchy product diverges by Proposition 3.1.1.

Subsection 3.3.2 Raabe’s Test

The ratio test is inconclusive when \(|x_{n+1}|/|x_n| \to 1\text{.}\) Raabe’s test refines that situation by measuring how quickly the ratio approaches \(1\text{.}\)

Proof.
We first prove part (1). By discarding finitely many terms, we may assume the displayed inequality holds for every \(n \ge 1\text{.}\) Then
\begin{equation*} (n+1)|x_{n+1}| \le (n+1)|x_n|-a|x_n| \end{equation*}
for every \(n \ge 1\text{.}\) Rearranging gives
\begin{equation*} 0 \lt (a-1)|x_n| \le n|x_n|-(n+1)|x_{n+1}|. \end{equation*}
Summing from \(n=1\) to \(n=N\text{,}\) we obtain
\begin{equation*} 0 \lt (a-1)\sum_{n=1}^N |x_n| \le |x_1|-(N+1)|x_{N+1}| \le |x_1|. \end{equation*}
Hence the partial sums of \(\sum |x_n|\) are bounded above. By Proposition 3.2.3, the series \(\sum |x_n|\) converges. Therefore \(\sum x_n\) converges absolutely.
Now consider part (2). Again we may discard finitely many terms and assume the inequality holds for all \(n \ge 1\text{.}\) Then
\begin{equation*} (n+1)|x_{n+1}| \ge (n+1)|x_n|-a|x_n| \ge n|x_n| \end{equation*}
for every \(n \ge 1\text{.}\) Thus the sequence \((n|x_n|)\) is increasing. Since \(x_n \ne 0\text{,}\) we have \(|x_1|>0\text{,}\) and therefore
\begin{equation*} |x_n| \ge \frac{|x_1|}{n} \end{equation*}
for every \(n \ge 1\text{.}\) Because the harmonic series diverges by Example 3.2.4, the comparison test shows that \(\sum |x_n|\) diverges. So \(\sum x_n\) is not absolutely convergent.

Proof.
Suppose \(a>1\text{.}\) Choose a number \(a_1\) with \(1 \lt a_1 \lt a\text{.}\) Then for all sufficiently large \(n\text{,}\)
\begin{equation*} (n+1)\left(1-\left|\frac{x_{n+1}}{x_n}\right|\right) \gt a_1. \end{equation*}
Rearranging, we obtain
\begin{equation*} \left|\frac{x_{n+1}}{x_n}\right| \lt 1-\frac{a_1}{n+1} \end{equation*}
eventually. Part (1) of Proposition 3.3.4 now shows that \(\sum x_n\) converges absolutely.
Suppose instead that \(a<1\text{.}\) Choose \(a_1\) with \(a \lt a_1 \lt 1\text{.}\) Then for all sufficiently large \(n\text{,}\)
\begin{equation*} (n+1)\left(1-\left|\frac{x_{n+1}}{x_n}\right|\right) \lt a_1, \end{equation*}
so
\begin{equation*} \left|\frac{x_{n+1}}{x_n}\right| \gt 1-\frac{a_1}{n+1} \end{equation*}
eventually. Part (2) of Proposition 3.3.4 implies that \(\sum x_n\) is not absolutely convergent.

Example 3.3.6.

Raabe’s test recovers the \(p\)-series criterion for \(p \ne 1\text{.}\) Let \(x_n=1/n^p\text{.}\) Then
\begin{equation*} \left|\frac{x_{n+1}}{x_n}\right| = \left(\frac{n}{n+1}\right)^p. \end{equation*}
A short computation shows
\begin{equation*} \lim (n+1)\left(1-\left(\frac{n}{n+1}\right)^p\right)=p. \end{equation*}
Therefore Corollary 3.3.5 gives absolute convergence when \(p>1\) and failure of absolute convergence when \(p<1\text{.}\) Since the terms are nonnegative, this means the series diverges when \(p<1\text{.}\) The case \(p=1\) is inconclusive for Raabe’s test.
References: compare [2], Chapter 3, for Cauchy products, and [9] for Raabe’s test and related refinements of the ratio test.
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