Proposition 3.2.1. Geometric Series.
Let \(r\in\R\text{.}\) The geometric series
\begin{equation*}
\sum_{n=0}^\infty r^n
\end{equation*}
converges to \(1/(1-r)\) when \(|r|<1\text{,}\) and diverges when \(|r|\ge 1\text{.}\)
Section 3.2 Convergence Tests for Series
In Section 3.1 we defined convergence of a series and proved the divergence test (Proposition 3.1.1). Here we develop positive tests that help us decide whether a series converges.
A series \(\sum x_n\) is called a Cauchy series if its sequence of partial sums is a Cauchy sequence. By Proposition 2.4.3 and Theorem 2.4.5, a series of real numbers converges if and only if it is Cauchy.
Subsection 3.2.1 Geometric Series
Proof.
Let \(s_k=\sum_{n=0}^k r^n\text{.}\) If \(r\ne 1\text{,}\) then
\begin{equation*}
(1-r)s_k = (1-r)(1+r+r^2+\cdots+r^k)=1-r^{k+1},
\end{equation*}
so
\begin{equation*}
s_k=\frac{1-r^{k+1}}{1-r}.
\end{equation*}
Suppose first that \(|r|<1\text{.}\) Then \(|r^n| = |r|^n
\to 0\) by Proposition 2.2.1 and so \((r^n)\) is null. Therefore, \(s_k\to 1/(1-r)\text{,}\) and the series converges to that value.
If \(|r| \ge 1\text{,}\) then \(|r^n|\ge 1\) for all \(n\text{,}\) so \(r^n\) is not a null sequence. Thus, the series diverges by Proposition 3.1.1.
Example 3.2.2.
Taking \(r=1/3\) gives
\begin{equation*}
1+\frac13+\frac19+\frac1{27}+\cdots
=
\frac{1}{1-\frac13}
=
\frac32.
\end{equation*}
Taking \(r=-1/2\) gives
\begin{equation*}
1-\frac12+\frac14-\frac18+\cdots
=
\frac{1}{1+\frac12}
=
\frac23.
\end{equation*}
Subsection 3.2.2 Series with Nonnegative Terms
For a series with nonnegative terms, the sequence of partial sums is automatically increasing. This gives a useful criterion.
Proposition 3.2.3. Nonnegative Term Criterion.
Suppose \(x_n\ge 0\) for all \(n\text{.}\) Then the series \(\sum x_n\) converges if and only if its sequence of partial sums is bounded above.
Proof.
Let \(s_k=\sum_{n=1}^k x_n\text{.}\) Since each \(x_n\ge 0\text{,}\) we have \(s_{k+1}-s_k=x_{k+1}\ge 0\text{,}\) so \((s_k)\) is increasing. An increasing sequence converges if and only if it is bounded above by Theorem 2.1.5. Since the series converges exactly when \((s_k)\) converges, the result follows.
Example 3.2.4.
The harmonic series
\begin{equation*}
\sum_{n=1}^\infty \frac{1}{n}
\end{equation*}
diverges.
Solution.
Group the terms as
\begin{equation*}
1+\frac12+\left(\frac13+\frac14\right)+
\left(\frac15+\frac16+\frac17+\frac18\right)+\cdots.
\end{equation*}
For each \(k\ge 1\text{,}\) the block from \(n=2^k+1\) to \(n=2^{k+1}\) has \(2^k\) terms, and each of those terms is at least \(1/2^{k+1}\text{.}\) Hence each block has sum at least \(1/2\text{.}\) Therefore the partial sums are unbounded above, so the series diverges by Proposition 3.2.3.
The previous example is the basic comparison for the \(p\)-series.
Proposition 3.2.5. \(p\)-Series Test.
For \(p\in\R\text{,}\) the series
\begin{equation*}
\sum_{n=1}^\infty \frac{1}{n^p}
\end{equation*}
converges if and only if \(p>1\text{.}\)
Proof.
If \(p\le 1\text{,}\) then \(1/n^p \ge 1/n\) for all \(n\in\N\text{.}\) Hence every partial sum of \(\sum 1/n^p\) is at least as large as the corresponding partial sum of the harmonic series. Since the harmonic series diverges by Example 3.2.4, the series \(\sum 1/n^p\) also diverges.
Now suppose \(p>1\text{.}\) For each \(k\ge 0\text{,}\) every \(n\) with \(2^k\le n \lt 2^{k+1}\) satisfies \(1/n^p \le 1/(2^k)^p\text{.}\) Therefore
\begin{equation*}
\sum_{n=2^k}^{2^{k+1}-1}\frac{1}{n^p}
\le
2^k\frac{1}{(2^k)^p}
=
2^{k(1-p)}.
\end{equation*}
Since \(0\lt 2^{1-p}\lt 1\text{,}\) the geometric series \(\sum_{k=0}^\infty 2^{k(1-p)}\) converges by Proposition 3.2.1. Hence the partial sums of \(\sum 1/n^p\) are bounded above, and so the series converges by Proposition 3.2.3.
For series with non-negative terms, convergence and absolute convergence mean the same thing. In general, absolute convergence implies convergence. The converse is false, since there are conditionally convergent series, for example the alternating series \(\sum (-1)^{n+1}
1/n\) (Example 3.2.16).
Proposition 3.2.6. Absolute Convergence Test.
If the series \(\sum |x_n|\) converges, then the series \(\sum x_n\) converges.
Proof.
Let \(u_k=\sum_{n=1}^k |x_n|\) and \(s_k=\sum_{n=1}^k x_n\text{.}\) Since \((u_k)\) converges, it is Cauchy by Proposition 2.4.3. If \(m>n\text{,}\) then
\begin{align*}
|s_m-s_n|
& =
|x_{n+1}+x_{n+2}+\cdots+x_m|\\
& \le
|x_{n+1}|+|x_{n+2}|+\cdots+|x_m| = u_m-u_n.
\end{align*}
Hence \((s_k)\) is also Cauchy. By Theorem 2.4.5, the sequence \((s_k)\) converges, so \(\sum x_n\) converges.
Subsection 3.2.3 Comparison Tests
The first comparison tool is the comparison test.
Proposition 3.2.7. Comparison Test.
Suppose \(0\le x_n\le y_n\) for all \(n\text{.}\) If \(\sum y_n\) converges, then \(\sum x_n\) converges.
Proof.
Let \(s_k=\sum_{n=1}^k x_n\) and \(t_k=\sum_{n=1}^k y_n\text{.}\) Then \(0\le s_k\le t_k\) for every \(k\text{.}\) If \(\sum y_n\) converges, then \((t_k)\) is bounded above. Consequently \((s_k)\) is bounded above as well, and Proposition 3.2.3 shows that \(\sum x_n\) converges.
A useful refinement is the limit comparison test.
Proposition 3.2.8. Limit Comparison Test.
Suppose \(x_n>0\) and \(y_n>0\) for all \(n\text{.}\) If
\begin{equation*}
\lim_{n\to\infty}\frac{x_n}{y_n}=L
\end{equation*}
for some \(L>0\text{,}\) then \(\sum x_n\) converges if and only if \(\sum y_n\) converges.
If instead
\begin{equation*}
\lim_{n\to\infty}\frac{x_n}{y_n}=0,
\end{equation*}
then convergence of \(\sum y_n\) implies convergence of \(\sum x_n\text{.}\)
Proof.
First suppose \(L>0\text{.}\) Then for all sufficiently large \(n\)
\begin{equation*}
\frac{L}{2} \lt \frac{x_n}{y_n} \lt \frac{3L}{2}
\end{equation*}
Since \(y_n>0\text{,}\) this is equivalent to
\begin{equation*}
\frac{L}{2}y_n \lt x_n \lt \frac{3L}{2}y_n.
\end{equation*}
If \(\sum y_n\) converges, then so does \(\sum (3L/2)y_n\text{,}\) and the comparison test gives convergence of \(\sum x_n\text{.}\) Conversely, if \(\sum x_n\) converges, then so does \(\sum (2/L)x_n\text{,}\) and the comparison test applied to \(y_n \lt (2/L)x_n \) shows that \(\sum y_n\) converges.
If \(x_n/y_n\to 0\text{,}\) then for all sufficiently large \(n\)\(x_n/y_n<1\text{.}\) Hence \(0\lt x_n\lt y_n\) eventually, so convergence of \(\sum y_n\) implies convergence of \(\sum
x_n\) by the comparison test.
Example 3.2.9.
The series
\begin{equation*}
\sum_{n=2}^\infty \frac{1}{n^3-1}
\end{equation*}
converges. Indeed,
\begin{equation*}
\lim_{n\to\infty}
\frac{\frac{1}{n^3-1}}{\frac{1}{n^3}}
=
\lim_{n\to\infty}\frac{n^3}{n^3-1}
=
1,
\end{equation*}
and \(\sum 1/n^3\) converges by Proposition 3.2.5. Therefore \(\sum 1/(n^3-1)\) converges by Proposition 3.2.8.
Subsection 3.2.4 Ratio and Root Tests
The next tests compare a series to a geometric series. The first is the ratio test.
Proposition 3.2.10. Ratio Test.
Let \((x_n)\) be a sequence of real numbers such that \(x_n\ne 0\) for all sufficiently large \(n\text{,}\) and suppose
\begin{equation*}
\lim_{n\to\infty}\frac{|x_{n+1}|}{|x_n|}=L.
\end{equation*}
If \(L<1\text{,}\) then \(\sum x_n\) converges absolutely. If \(L>1\text{,}\) then \(\sum x_n\) diverges.
Proof.
Suppose \(L<1\text{.}\) Then \(L\lt r:=(1+L)/2 \lt 1\text{,}\) so for all \(n\) sufficiently large,
\begin{equation*}
\frac{|x_{n+1}|}{|x_n|}\lt r.
\end{equation*}
By dropping finitely many terms, which does not affect convergence, we can assume the inequality holds for all \(n\text{.}\) Hence,
\begin{gather*}
|x_2| < |x_1|\\
|x_3| < r|x_2| < r^2|x_1|\\
\vdots\\
|x_n| < r^{n-1}|x_1|.
\end{gather*}
Since \(0 \lt r \lt 1\text{,}\) \(\sum r^k\) converges by Proposition 3.2.1 and the comparison test implies that \(\sum_{n=N}^\infty |x_n|\) converges. Therefore \(\sum x_n\) converges absolutely.
Now suppose \(L>1\text{,}\) and set \(r=(1+L)/2\text{.}\) Then \(1\lt r\lt L\text{,}\) so there exists \(N\) such that \(|x_{n+1}|/|x_n| \gt r\gt 1\) for all \(n\ge N\text{.}\) In particular, \(|x_n|\) is eventually strictly increasing, so it cannot converge to \(0\text{.}\) Therefore \(\sum x_n\) diverges by Proposition 3.1.1.
Example 3.2.11.
The series
\begin{equation*}
\sum_{n=1}^\infty \frac{n^2}{2^n}
\end{equation*}
converges absolutely, because
\begin{equation*}
\frac{(n+1)^2/2^{n+1}}{n^2/2^n}
=
\frac{(n+1)^2}{2n^2}
\to
\frac12.
\end{equation*}
If the ratio limit is \(1\text{,}\) the ratio test gives no conclusion: both \(\sum 1/n\) and \(\sum 1/n^2\) have consecutive-term ratio tending to \(1\text{,}\) but the first diverges and the second converges.
The root test provides another way of comparing a series with a geometric series. It is often more convenient when an explicit \(n\)-th power appears in the terms.
Proposition 3.2.12. Root Test.
Let
\begin{equation*}
L=\limsup_{n\to\infty}|x_n|^{1/n}.
\end{equation*}
If \(L<1\text{,}\) then \(\sum x_n\) converges absolutely. If \(L>1\text{,}\) then \(\sum x_n\) diverges.
Proof.
Suppose \(L<1\text{.}\) Then for some \(N\text{,}\)
\begin{equation*}
\sup_{n \ge N} |x_n|^{1/n} < r := \frac{1+L}{2} < 1.
\end{equation*}
That means \(|x_n| < r^n\) for all \(n \ge N\text{.}\) Since \(0 < r < 1\text{,}\) the geometric series \(\sum r^n\) converges, and by the comparison test so does \(\sum |x_n|\text{.}\)
Now suppose \(L>1\text{,}\) and let \(r := (1+L)/2\text{.}\) Then \(r>1\text{.}\) By the definition of limit superior, for every \(N\text{,}\)
\begin{equation*}
\sup_{n \ge N} |x_n|^{1/n} \ge r.
\end{equation*}
Hence for every \(N\) there exists \(n \ge N\) such that \(|x_n|^{1/n} \ge r\text{.}\) Therefore
\begin{equation*}
|x_n| \ge r^n \ge 1
\end{equation*}
for infinitely many \(n\text{.}\) As a result \((x_n)\) cannot be a null sequence, and \(\sum x_n\) diverges by Proposition 3.1.1.
Example 3.2.13.
The series
\begin{equation*}
\sum_{n=1}^\infty \left(\frac{3n}{4n+1}\right)^n
\end{equation*}
converges absolutely, because
\begin{equation*}
\left|\left(\frac{3n}{4n+1}\right)^n\right|^{1/n}
=
\frac{3n}{4n+1}
\to
\frac34
<
1.
\end{equation*}
The root test is also inconclusive when \(L=1\text{.}\) For example, \(\sum 1/n\) diverges while \(\sum 1/n^2\) converges, but both satisfy \(\lim |x_n|^{1/n}=1\text{.}\)
By Theorem 2.3.4, if the limit \(\lim |x_{n+1}|/|x_n|\) exists, then
\begin{equation*}
\limsup |x_n|^{1/n} = \lim \frac{|x_{n+1}|}{|x_n|}.
\end{equation*}
Therefore the root test is at least as strong as the ratio test. The following example shows that it is strictly stronger.
Example 3.2.14.
The root test can succeed even when the ratio test fails. Consider the series
\begin{equation*}
\sum_{n=1}^{\infty} x_n,
\end{equation*}
where
\begin{equation*}
x_n =
\begin{cases}
2^{-n}, & \text{if } n \text{ is even},\\
3^{-n}, & \text{if } n \text{ is odd}.
\end{cases}
\end{equation*}
Then
\begin{equation*}
|x_n|^{1/n} =
\begin{cases}
1/2, & \text{if } n \text{ is even},\\
1/3, & \text{if } n \text{ is odd},
\end{cases}
\end{equation*}
so \(\limsup |x_n|^{1/n} = 1/2 < 1\text{.}\) Hence the root test shows that \(\sum x_n\) converges absolutely.
On the other hand,
\begin{equation*}
\frac{x_{n+1}}{x_n} =
\begin{cases}
2^n/3^{n+1}, & \text{if } n \text{ is even},\\
3^n/2^{n+1}, & \text{if } n \text{ is odd},
\end{cases}
\end{equation*}
so one subsequence tends to \(0\) and the other tends to \(+\infty\text{.}\) Therefore \((x_{n+1}/x_n)\) has no limit, and the ratio test does not apply.
Subsection 3.2.5 Alternating Series
The final test in this section is the alternating series test.
Proposition 3.2.15. Alternating Series Test.
Suppose \(x_n\ge 0\) for all \(n\text{,}\) the sequence \((x_n)\) is decreasing, and \(x_n\to 0\text{.}\) Then the alternating series
\begin{equation*}
\sum_{n=1}^\infty (-1)^{n+1}x_n
\end{equation*}
converges.
Proof.
Let \(s_k=\sum_{n=1}^k (-1)^{n+1}x_n\text{.}\) Then
\begin{equation*}
s_{2k+2}-s_{2k}=x_{2k+1}-x_{2k+2}\ge 0,
\end{equation*}
so the even partial sums form an increasing sequence. Also,
\begin{equation*}
s_{2k+3}-s_{2k+1}=-x_{2k+2}+x_{2k+3}\le 0,
\end{equation*}
so the odd partial sums form a decreasing sequence.
For every \(k\text{,}\) we have \(s_{2k}\le s_{2k+1}\le s_1=x_1\text{.}\) Thus \((s_{2k})\) is increasing and bounded above, so it converges by Theorem 2.1.5. Write \(s_{2k}\to L\text{.}\) Since
\begin{equation*}
s_{2k+1}-s_{2k}=x_{2k+1}\to 0,
\end{equation*}
the limit laws imply that \(s_{2k+1}\to L\) as well. Therefore the whole sequence of partial sums converges to \(L\text{,}\) and the series converges.
Example 3.2.16.
The alternating harmonic series
\begin{equation*}
\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}
\end{equation*}
converges by Proposition 3.2.15, because \((1/n)\) is decreasing and null. It is not absolutely convergent, since
\begin{equation*}
\sum_{n=1}^\infty \left|\frac{(-1)^{n+1}}{n}\right|
=
\sum_{n=1}^\infty \frac{1}{n}
\end{equation*}
diverges. Hence it is conditionally convergent.
The proof shows more: if \(L=\sum_{n=1}^\infty (-1)^{n+1}x_n\text{,}\) then the error after \(n\) terms satisfies
\begin{equation*}
|L-s_n|\le x_{n+1}.
\end{equation*}
So the first omitted term controls the truncation error.
