1. Series Introduction
Important
Prior to watching this video, you should read about Cauchy sequences. The key facts about Cauchy sequences you'll need to know are the definition and that a sequence converges if and only if it is a Cauchy sequence.
1.1. Series Defined
Suppose that \(\{a_n\}_{n=m}^\infty\) is an infinite sequence of real numbers. We say that the infinite series
\[ \sum_{n=m}^\infty a_n \]
is convergent when the limit of the partial sums exists, namely,
\[ \lim_{N \rightarrow \infty} \sum_{n=m}^N a_n \]
exists, and in that case we define
\[ \sum_{n=m}^\infty a_n := \lim_{N \rightarrow \infty} \sum_{n=m}^N a_n. \]
Proposition (Nth Term Test)
If \(\{a_n\}_{n \in {\mathbb N}}\) is a sequence of real numbers and the series
\[ \sum_{n=1}^\infty a_n \]
converges, then \(\lim_{m \rightarrow \infty} a_m = 0\).
Proof
Because the partial sums \(\sum_{n=1}^N a_n\) are a convergent sequence, they must be a Cauchy sequence. Therefore for any \(\epsilon\), there is an index \(M\) such that \(|\sum_{n=1}^{N_1} a_n - \sum_{n=1}^{N_2} a_n| < \epsilon\) whenever \(N_1\) and \(N_2\) are greater than or equal to \(M\). In particular, if \(m \geq M+1\), then
\[ |a_m| = \left| \sum_{n=1}^{m} a_n - \sum_{n=1}^{m-1} a_n \right| < \epsilon, \]
which gives that \(a_m \rightarrow 0\) as \(m \rightarrow \infty\).
Important (The Harmonic Series)
The harmonic series \(\sum_{n=1}^\infty \frac{1}{n}\) is not convergent despite the fact that the terms tend to zero. In fact,
\[ \sum_{n=1}^\infty \frac{1}{n} = \infty. \]
Proof
If the harmonic series were convergent, then the partial sums
\[ S_N := \sum_{n=1}^N \frac{1}{n}\]
would themselves form a Cauchy sequence. But for any natural number \(m\), we have the inequality
\[{}S_{2^{m+1}} - S_{2^m}{}\]
\[{}= \sum_{n=2^{m}+1}^{2^{m+1}} \frac{1}{n}{}\]
\[{}\geq \sum_{n=2^m+1}^{2^{m+1}} 2^{-(m+1)}{}\]
\[{}\geq \frac{1}{2}. {}\]
Since \(2^m \rightarrow \infty\), it cannot be the case that \(|S_{N_1} - S_{N_2}| < \frac{1}{2}\) for all \(N_1\) and \(N_2\) sufficiently large. Since the partial sums are a monotone sequence that does not converge, they must be unbounded, i.e., \(\sum_{n=1}^N \frac{1}{N} \rightarrow \infty\) as \(N \rightarrow \infty\).
Example
If \(\rho\) is any fixed real number, then the series
\[ \sum_{n=0}^\infty \rho^n \]
converges if and only if \(|\rho| < 1\). This follows rather immediately from the convenient formula one can prove (by induction) for the partial sums:
\[ \sum_{n=0}^{N} \rho^n = \begin{cases} \frac{1-\rho^{N+1}}{1-\rho} & \rho \neq 1, \\ N+1 & \rho = 1. \end{cases}\]
2. Convergence Tests
Some of the standard convergence tests are proved below. Remember: These tests only scratch the surface and there will be many situations in which none of them apply. Focus not just on what the tests are and how they are used, but on the techniques which go into their proofs, as the underlying techniques are much more widely applicable than the individual tests.
2.1. Absolute Convergence Test
Theorem
If the series
\[ \sum_{n=1}^\infty |a_n| \]
converges, then the series
\[ \sum_{n=1}^\infty a_n \]
converges.
Proof
Important (Conditional Convergence)
The converse is not true: there are plenty of series such that
\[ \sum_{n=1}^\infty a_n \]
converges while
\[ \sum_{n=1}^\infty |a_n| \]
does not. These are called conditionally convergent series. Any series which converges when its terms are replaced by their absolute values is called absolutely convergent.
2.2. Direct Comparison Test
Theorem
Suppose that \(\{a_n\}_{n=1}^\infty\) and \(\{b_n\}_{n=1}^\infty\) are sequences such that \(|a_n| \leq |b_n|\) for each \(n\). If
\[ \sum_{n=1}^\infty |b_n| \]
is a convergent series, then so are
\[ \sum_{n=1}^\infty |a_n| \text{ and } \sum_{n=1}^\infty a_n. \]
Proof
2.3. Ratio and Root Tests
Theorem
Suppose \(\{a_n\}_{n=1}^\infty\) is a sequence of nonzero real numbers. If
\[ \limsup_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \]
or
\[ \limsup_{n \rightarrow \infty} |a_n|^{1/n} < 1 \]
then the series \(\sum_{n=1}^\infty a_n\) converges absolutely. (Recall that when the limit itself exists, it agrees with the limsup.)
Proof
In the first case,
\[ \limsup_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \]
means that there is some \(N\) and som
\[ \inf_N \sup_{n \geq N} \left| \frac{a_{n+1}}{a_n} \right| = \rho_0 < 1. \]
As the infimun is strictly less than \(1\), there must actually be some value of \(N\) and some \(\rho < 1\) such that
\[ \sup_{n \geq N} \left| \frac{a_{n+1}}{a_n} \right| \leq \rho < 1. \]
This means that
\[{}\left| \frac{a_{N+k}}{a_N} \right|{}\]
\[{}= \left| \frac{a_{N+1}}{a_N} \frac{a_{N+2}}{a_{N+1}} \cdots \frac{a_{N+k}}{a_{N+k-1}} \right|{}\]
\[{}\leq \rho^k.{}\]
In other words, \(|a_n| \leq \rho^{-N} |a_N| \rho^n\) for all \(n \geq N\). Fix \(C := \rho^{-N} |a_N|\). Let \(b_n\) be the sequence such that \(b_n := a_n\) for \(n < N\) and \(b_n := C \rho^{n}\) for \(n \geq N\). This series satisfies \(|a_n| \leq |b_n|\) for each \(n\). Because it is eventually equal to a constant multiple of a convergent geometric series, \(\sum_{n=1}^\infty |b_n|\) converges. By the Direct Comparison Test, \(\sum_{n=1}^\infty |a_n|\) must also converge.
In the root test scenario, one can use the fact that
\[ \limsup_{n \rightarrow \infty} |a_n|^{1/n} = \rho_0 < 1 \]
to establish that there is some \(N\) and some \(\rho < 1\) such that
\[ \sup_{n \geq N} |a_n|^{1/n} = \rho < 1 \]
to once again see that there is domination by a geometric series: \(|a_n| \leq \rho^n\) for all \(n \geq N\).
As a side note, by reverse-engineering the proof above, it's not so hard to see that the root test always applies at least as often as the ratio test. So in a technical sense, the root test is strictly stronger than the ratio test. In practice, there are cases when any one of the two is easier to verify than the other.
2.4. Alternating Series Test
Theorem
If \(\{a_n\}_{n=1}^\infty\) is a sequence of nonnegative terms which decreases to zero, then
\[ \sum_{n=1}^\infty (-1)^{n-1} a_n \]
converges and moreover
\[ 0 \leq \sum_{n=1}^\infty (-1)^{n-1} a_n \leq a_1. \]
Proof
The key is to observe that for any positive decreasing sequence of numbers \(b_1 \geq b_2 \geq \cdots \geq b_k\),
\[{}0{}\]
\[{}\leq b_1 - b_2 + b_3 + \cdots{}\]
\[{}+ (-1)^{k-1} b_k \leq b_1.{}\]
This is easily proved by induction on \(k\). When the number of terms is \(1\), the inequality is trivial: \(0 \leq b_1 \leq b_1\) simply because the terms are assumed to be positive. For two terms, \(0 \leq b_1 - b_2 \leq b_1\) is basically just a restatement of the assumptions that the terms are decreasing (so \(b_1 - b_2 \geq 0\)) and positive (so \(b_1 - b_2 \leq b_1\)). By induction, for larger values of \(k\), we can take
\[ b_1 - b_2 + b_3 + \cdots + (-1)^{k-1} b_k \]
and write it as
\[ (b_1 - b_2) + (b_3 + \cdots + (-1)^{k-1} b_k) \]
and then use the induction hypothesis to bound the second batch of terms in parentheses:
\[ 0 \leq b_3 + \cdots + (-1)^{k-1} b_k \leq b_3 \]
so
\[{}b_1 - b_2{}\]
\[{}\leq b_1 - b_2 + b_3 + \cdots{}\]
\[{}+ (-1)^{k-1} b_k \leq b_1 - b_2 + b_3.{}\]
Because the terms are decreasing, \(b_1 - b_2 \geq 0\) and \(b_1 - b_2 + b_3 \leq b_1\) as desired.
Now our observation does all the work. First off, it shows that all partial sums of the series \(\sum_{n=1}^\infty (-1)^{n-1} a_n\) are between \(0\) and \(a_1\). This means that the limit (if it exists) will also be between \(0\) and \(a_1\). If we consider the difference of the \(N\)-th partial sum and the \(M\)-th partial sum (assuming without loss of generality that \(M < N\),)
\[{}\sum_{n=1}^N (-1)^{n-1} a_n {}\]
\[{}- \sum_{n=1}^M (-1)^{n-1} a_n{}\]
\[{}= \sum_{n=M+1}^N (-1)^{n-1} a_n. {}\]
If we use our observation, it tells us that the magnitude of this difference is never greater than \(|a_{M+1}|\). Since \(a_{n}\) decreases to zero, if both \(N\) and \(M\) are sufficiently large, the magnitude of the difference will necessarily be less than \(\epsilon\).
Exercises
- Give an explicit example of a series which is conditionally convergent and explain both why it must be convergent and why it cannot be absolutely convergent.
- Show that the series\[ \sum_{n=0}^\infty \frac{1}{n!}\]converges. Then prove that \(n! \geq 2^{n-1}\) for all \(n \geq 2\) and use this to show that the sum of the series in question is no greater than 3.
- Give an example of a sequence of terms \(\{a_n\}_{n=1}^\infty\) which are nonnegative and tend to zero but violate the conclusion of the alternating series test.