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where the elements \(a_{n}\) come from a number system in which addition is defined. We will fix our number system to be the set of complex numbers \(\mathbb{C}\), but we give ourselves the freedom of restricting to the real numbers or even the rational numbers.
An infinite series \(\sum_{k=1}^{\infty}a_{k}\) is said to _converge_ to a _sum_\(s\) if its sequence of _partial sums_\(s_{n}=\sum_{k=1}^{n}a_{k}\) converges to \(s\). We write
\[\sum_{k=1}^{\infty}a_{k}=s.\]
If \(\{s_{n}\}\) diverges, the series is said to diverge. Note that every sequence \(\{a_{n}\}\) has a representation as the partial sum of a series, putting \(s_{1}=a_{1}\) and \(a_{n}=s_{n}-s_{n-1}\) and \(s_{n}=\sum_{k=1}^{n}a_{k}\), and vice versa. Therefore, in a certain sense sequences and series are in one-to-one correspondence.
**Example 28**.: The infinite series \(\sum_{k=1}^{\infty}\dfrac{1}{k(k+1)}\) converges to \(1\). We see this by finding its partial sums:
\[s_{n}=\sum_{k=1}^{n}\dfrac{1}{k(k+1)}=\sum_{k=1}^{n}\left(\dfrac{1}{k}-\dfrac {1}{k+1}\right)=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}.\]Thus, \(\sum_{k=1}^{\infty}\frac{1}{k(k+1)}=\lim_{n\to\infty}s_{n}=\lim_{n\to\infty}\frac{n} {n+1}=1\).
However, most series do not have partial sum sequences with a nice simple formula like the one in the above example. There are two questions we need to answer about series:
* How do we tell if a given series converges or diverges?
* If a series does converge, how do we find its sum?
Although it requires certain work to decide if a given series converges or diverges, the second question presents much more difficulties than the first one. We now start by applying some well-known theorems about sequences to the sequence of partial sums, to obtain some useful criteria.
**Theorem 19**.: _A series with nonnegative terms converges if and only if its sequence of partial sums forms a bounded sequence._
This follows from the fact that for a series with nonnegative terms \(a_{k}\geq 0\) for all \(k\), \(s_{n}\leq s_{n+1}\). If \(\{s_{n}\}\) is monotonic, then \(\{s_{n}\}\) converges if and only if it is bounded.
**Theorem 20**.: _A series \(\sum_{k=1}^{\infty}a_{k}\) converges if and only if for every \(\epsilon>0\) there is an integer \(N\) such that \(\left|\sum_{k=n}^{m}a_{k}\right|\leq\epsilon\) for \(m\geq n\geq N\)._
This is a direct consequence of the Cauchy criterion for sequences. In particular, if \(m=n\), then the conclusion of the above theorem becomes \(|a_{n}|\leq\epsilon\) whenever \(n\geq N\), which means that if \(\sum_{n=1}^{\infty}a_{n}\) converges, then \(\lim_{n\to\infty}a_{n}=0\). Thus we can conclude if \(\lim_{n\to\infty}a_{n}\neq 0\), then \(\sum_{n=1}^{\infty}a_{n}\) diverges. This is called the _divergence test_. However the condition \(\lim_{n\to\infty}a_{n}=0\) does not ensure the convergence of \(\sum_{n=1}^{\infty}a_{n}\). A typical counterexample is the series \(\sum_{n=1}^{\infty}\frac{1}{n}\) which we will see diverges (see Example 30 below). Another very important example of a series is the following _geometric series_:
**Example 29**.: If \(x\neq 1\), the geometric series \(\sum_{k=0}^{\infty}x^{k}\) has partial sums
\[s_{n}=\sum_{k=0}^{n}x^{k}=\frac{1-x^{n+1}}{1-x}.\]This can be proved using mathematical induction. The geometric series converges if and only if \(|x|<1\), in which case the sum is \(s=\dfrac{1}{1-x}\). This is because \(\lim_{n\to\infty}x^{n}=0\) for \(|x|<1\) and thus
\[\lim_{n\to\infty}s_{n}=\lim_{n\to\infty}\dfrac{1-x^{n+1}}{1-x}=\dfrac{1}{1-x}.\]
For \(x=1\), we get
\[\sum_{k=0}^{\infty}x^{k}=1+1+1+\cdots,\]
which clearly diverges.
**Example 30** (Harmonic Series).: The series \(\sum_{n=1}^{\infty}\dfrac{1}{n}\) is called the _harmonic series._
Clearly its sequence of partial sums
\[s_{m}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{m}\]
is increasing, but it is increasing at a slow pace, so we might think naively that this sequence of partial sums to be somewhat bounded. However,
\[s_{4}=1+\dfrac{1}{2}+\left(\dfrac{1}{3}+\dfrac{1}{4}\right)>1+\dfrac{1}{2}+ \left(\dfrac{1}{4}+\dfrac{1}{4}\right)=2,\]
and a similar calculation yields \(s_{8}>1+3\left(\frac{1}{2}\right)\). Indeed, \(s_{2^{k}}\) is unbounded since
\[s_{2^{k}}=1+\dfrac{1}{2}+\left(\dfrac{1}{3}+\dfrac{1}{4}\right)+\left(\dfrac{ 1}{5}+\cdots+\dfrac{1}{8}\right)+\cdots+\left(\dfrac{1}{2^{k-1}+1}+\cdots+ \dfrac{1}{2^{k}}\right)\]
\[>1+\dfrac{1}{2}+\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{1}{8}+ \cdots+\dfrac{1}{8}\right)+\cdots+\left(\dfrac{1}{2^{k}}+\cdots+\dfrac{1}{2^{ k}}\right)\]
\[=1+\dfrac{1}{2}+2\left(\dfrac{1}{4}\right)+4\left(\dfrac{1}{8}\right)+\cdots +2^{k-1}\left(\dfrac{1}{2^{k}}\right)=1+k\left(\dfrac{1}{2}\right).\]
Thus the sequence of partial sums associated with the harmonic series is not bounded. Because convergent sequences are bounded, the harmonic series diverges.
The following theorem called the _Cauchy Condensation Test_ is remarkable in the sense that it gives convergence or divergence of \(\sum_{n=0}^{\infty}a_{n}\) by looking at a subsequence \(\{a_{2^{k}}\}\) of \(\{a_{k}\}\).
Proof
: Since the series has nonnegative terms, it converges if and only if its partial sums form a bounded sequence. Thus it suffices to show boundedness of the partial sums. First assume that \(\sum_{n=0}^{\infty}2^{n}a_{2^{n}}\) converges. Then the partial sums
\[t_{k}=a_{1}+2a_{2}+\cdots+2^{k}a_{2^{k}}\]
are bounded; that is, there exists a positive \(M>0\) such that \(t_{k}\leq M\) for all \(k\in\mathbb{N}\). Now we need to show that the partial sums
\[s_{m}=a_{1}+a_{2}+\cdots+a_{m}\]
are bounded. For \(m<2^{k}\),
\[s_{m}=a_{1}+(a_{2}+a_{3})+(a_{4}+a_{5}+a_{6}+a_{7})+\cdots+(a_{2^{k}}+\cdots+a _{2^{k+1}-1}).\]
Therefore,
\[s_{m}\leq a_{1}+(a_{2}+a_{2})+(a_{4}+a_{4}+a_{4}+a_{4})+\cdots+(a_{2^{k}}+ \cdots+a_{2^{k}})=t_{k}\leq M.\]
For the proof of reverse implication, consider \(m>2^{k}\), and show that \(2s_{m}\geq t_{k}\) to conclude that the sequences \(\{s_{m}\}\) and \(\{t_{k}\}\) are either both unbounded or both bounded.
**Corollary 4**.: _The series \(\sum_{n=1}^{\infty}\frac{1}{n^{p}}\) converges if \(p>1\) and diverges if \(p\leq 1\)._
Proof
: From the well-known fact that if \(\sum_{n=1}^{\infty}a_{n}\) converges, then \(\lim_{n\to\infty}a_{n}=0\), the given series diverges when \(p\leq 0\). Now, if \(p>0\), then we can use the Cauchy condensation test to examine the series \(\sum_{k=0}^{\infty}2^{k}\frac{1}{2^{kp}}=\sum_{k=0}^{\infty}2^{(1-p)k}\). Comparing with the geometric series by taking \(x=2^{1-p}\), we conclude that when \(2^{1-p}<1\), i.e., when \(1-p<0\), the series converges.
**Definition 17**.: A series \(\sum_{k=0}^{\infty}a_{k}\) is said to be _absolutely convergent_ if \(\sum_{k=0}^{\infty}|a_{k}|\) is convergent.
**Theorem 22**.: _Every absolutely convergent series converges._
Proof
: The proof follows from the Cauchy criterion for convergence and the inequality
\[|s_{m}-s_{n}|\leq\left|\sum_{k=n}^{m}a_{k}\right|\leq\sum_{k=n}^{m}|a_{k}|=t_ {m}-t_{n}.\qed\]Note that the above theorem is essentially equivalent to saying if \(|a_{k}|\leq b_{k}\) and if \(\sum_{k=0}^{\infty}b_{k}\) is convergent, \(\sum_{k=0}^{\infty}a_{k}\) is absolutely convergent.
**Remark 14**.: If \(\sum_{k=0}^{\infty}a_{k}\) converges, but \(\sum_{k=0}^{\infty}|a_{k}|\) diverges, we say \(\sum_{k=0}^{\infty}a_{k}\) is _conditionally convergent_ or \(\sum_{k=0}^{\infty}a_{k}\) converges _non-absolutely_. A well-known example of a conditionally convergent series is the _alternating harmonic series_\(\sum_{k=0}^{\infty}(-1)^{k}\dfrac{1}{k}\).
In the following we list the comparison, ratio, and root tests. They are tests for absolute convergence, and they cannot give information about conditionally convergent series. The basic theorem on series with alternating signs is due to G. Leibniz. For Leibniz’s alternating series theorem, we refer the reader to [18], pp. 72-73.
**Theorem 23** (Comparison Test).: _Suppose \(a_{n}>0\) for all \(n\in\mathbb{N}\) and \(\sum_{n=0}^{\infty}a_{n}\) converges. If \(b_{n}\in\mathbb{C}\) is such that_
\[|b_{n}|\leq a_{n}\qquad\text{for all }\,n,\]
_then the series \(\sum_{n=0}^{\infty}b_{n}\) converges absolutely and hence converges._
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