2.2: Infinite Series
- Page ID
- 22644
Given a sequence \(\left\{a_{i}\right\}_{i=m}^{\infty},\) let \(\left\{s_{n}\right\}_{n=m}^{\infty}\) be the sequence defined by
\[s_{n}=\sum_{i=m}^{n} a_{i}.\]
We call the sequence \(\left\{s_{n}\right\}_{n=m}^{\infty}\) an infinite series. If \(\left\{s_{n}\right\}_{n=m}^{\infty}\) converges, we call
\[s=\lim _{n \rightarrow \infty} s_{n}\]
the sum of the series. For any integer \(n \geq m,\) we call \(s_{n}\) a partial sum of the series.
We will use the notation
\[\sum_{i=1}^{\infty} a_{i}\]
to denote either \(\left\{s_{n}\right\}_{n=m}^{\infty},\) the infinite series, or \(s\), the sum of the infinite series. Of course, if \(\left\{s_{n}\right\}_{n=m}^{\infty}\) diverges, then we say \(\sum_{i=m}^{\infty} a_{i}\) diverges.
Suppose \(\sum_{i=m}^{\infty} a_{i}\) converges and \(\beta \in \mathbb{R} .\) Show that \(\sum_{i=m}^{\infty} \beta a_{i}\) also converges and
\[\sum_{i=m}^{\infty} \beta a_{i}=\beta \sum_{i=m}^{\infty} a_{i}.\]
Suppose both \(\sum_{i=m}^{\infty} a_{i}\) and \(\sum_{i=m}^{\infty} b_{i}\) converge. Show that \(\sum_{i=m}^{\infty}\left(a_{i}+b_{i}\right)\) converges and
\[\sum_{i=m}^{\infty}\left(a_{i}+b_{i}\right)=\sum_{i=m}^{\infty} a_{i}+\sum_{i=m}^{\infty} b_{i}.\]
Given an infinite series \(\sum_{i=m}^{\infty} a_{i}\) and an integer \(k \geq m,\) show that \(\sum_{i=m}^{\infty} a_{i}\) converges if and only if \(\sum_{i=k}^{\infty} a_{i}\) converges.
Suppose \(\sum_{i=m}^{\infty} a_{i}\) converges. Then \(\lim _{n \rightarrow \infty} a_{n}=0\).
- Proof
-
Let \(s_{n}=\sum_{i=m}^{n} a_{i}\) and \(s=\lim _{n \rightarrow \infty} s_{n} .\) Since \(a_{n}=s_{n}-s_{n-1},\) we have \(\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty}\left(s_{n}-s_{n-1}\right)=\lim _{n \rightarrow \infty} s_{n}-\lim _{n \rightarrow \infty} s_{n-1}=s-s=0\). Q.E.D.
Let \(s=\sum_{i=0}^{\infty}(-1)^{n} .\) Note that
\[s=\sum_{n=0}^{\infty}(-1)^{n}=1-\sum_{n=0}^{\infty}(-1)^{n}=1-s, \]
from which it follows that \(s=\frac{1}{2} .\) Is this correct?
Show that for any real number \(x \neq 1\),
\[s_{n}=\sum_{i=0}^{n} x^{i}=\frac{1-x^{n+1}}{1-x} .\]
(Hint: Note that \(\left.x^{n+1}=s_{n+1}-s_{n}=1+x s_{n}-s_{n} .\right)\)
For any real number \(x\) with \(|x|<1\),
\[\sum_{n=0}^{\infty} x^{n}=\frac{1}{1-x}.\]
- Proof
-
If \(s_{n}=\sum_{i=0}^{n} x^{i},\) then, by the previous exercise,
\[s_{n}=\frac{1-x^{n+1}}{1-x}.\]
Hence
\[\sum_{n=0}^{\infty} x^{n}=\lim _{n \rightarrow \infty} s_{n}=\lim _{n \rightarrow \infty} \frac{1-x^{n+1}}{1-x}=\frac{1}{1-x}.\]
2.2.1 Comparison Tests
The following two propositions are together referred to as the comparison test.
Suppose \(\sum_{i=m}^{\infty} a_{i}\) and \(\sum_{i=k}^{\infty} b_{i}\) are infinite series for which there exists an integer \(N\) such that \(0 \leq a_{i} \leq b_{i}\) whenever \(i \geq N .\) If \(\sum_{i=k}^{\infty} b_{i}\) converges, then \(\sum_{i=m}^{\infty} a_{i}\) converges.
- Proof
-
By Exercise 2.2 .3 We need only show that \(\sum_{i=N}^{\infty} a_{i}\) converges. Let \(s_{n}\) be the nth partial sum of \(\sum_{i=N}^{\infty} a_{i}\) and let \(t_{n}\) be the nth partial sum of \(\sum_{i=N}^{\infty} b_{i} .\) Now
\[s_{n+1}-s_{n}=a_{n+1} \geq 0\]
for every \(n \geq N,\) so \(\left\{s_{n}\right\}_{n=N}^{\infty}\) is a nondecreasing sequence. Moreover,
\[s_{n} \leq t_{n} \leq \sum_{i=N}^{\infty} b_{i}<+\infty\]
for every \(n \geq N .\) Thus \(\left\{s_{n}\right\}_{n=N}^{\infty}\) is a nondecreasing, bounded sequence, and so converges.
Suppose \(\sum_{i=m}^{\infty} a_{i}\) and \(\sum_{i=k}^{\infty} b_{i}\) are infinite series for which there exists an integer \(N\) such that \(0 \leq a_{i} \leq b_{i}\) whenever \(i \geq N .\) If \(\sum_{i=k}^{\infty} a_{i}\) diverges then \(\sum_{i=m}^{\infty} b_{i}\) diverges.
- Proof
-
By Exercise 2.2 .3 we need only show that \(\sum_{i=N}^{\infty} b_{i}\) diverges. Let \(s_{n}\) be the nth partial sum of \(\sum_{i=N}^{\infty} a_{i}\) and let \(t_{n}\) be the nth partial sum of \(\sum_{i=N}^{\infty} b_{i} .\) Now \(\left\{s_{n}\right\}_{n=N}^{\infty}, N\) is a nondecreasing sequence which diverges, and so we must have \(\lim _{n \rightarrow \infty} s_{n}=+\infty .\) Thus given any real number \(M\) there exists an integer \(L\) such that
\[M<s_{n} \leq t_{n}\]
whenever \(n>L .\) Hence \(\lim _{n \rightarrow \infty} t_{n}=+\infty\) and \(\sum_{i=m}^{\infty} b_{i}\) diverges. (\quad\) Q.E.D.
Consider the infinite series
\[\sum_{n=0}^{\infty} \frac{1}{n !}=1+1+\frac{1}{2}+\frac{1}{3 !}+\frac{1}{4 !}+\cdots .\]
Now for \(n=1,2,3, \dots,\) we have
\[0<\frac{1}{n !} \leq \frac{1}{2^{n-1}}.\]
Since
\[\sum_{n=1}^{\infty} \frac{1}{2^{n-1}}\]
converges, it follows that
\[\sum_{n=0}^{\infty} \frac{1}{n !}\]
converges. Moreover,
\[2<\sum_{n=0}^{\infty} \frac{1}{n !}<1+\sum_{n=1}^{\infty} \frac{1}{2^{n-1}}=1+\frac{1}{1-\frac{1}{2}}=3.\]
We let
\[e=\sum_{n=0}^{\infty} \frac{1}{n !}.\]
\(e \notin \mathbb{Q}\).
- Proof
-
Suppose \(e=\frac{p}{q}\) where \(p, q \in \mathbb{Z}^{+} .\) Let
\[a=q !\left(e-\sum_{i=0}^{q} \frac{1}{n !}\right).\]
Then \(a \in \mathbb{Z}^{+}\) since \(q! e=(q-1) ! p\) and \(n !\) divides \(q !\) when \(n \leq q .\) At the same time
\[\begin{aligned} a &=q ! \left(\sum_{n=0}^{\infty} \frac{1}{n !}-\sum_{i=0}^{q} \frac{1}{n !}\right) \\ &=q ! \sum_{n=q+1}^{\infty} \frac{1}{n !} \\ &=\left(\frac{1}{q+1}+\frac{1}{(q+1)(q+2)}+\frac{1}{(q+1)(q+2)(q+3)}+\cdots \right) \\ &= \frac{1}{q+1} \left(1+\frac{1}{q+2}+\frac{1}{(q+2)(q+3)}+\cdots\right) \\ &=\frac{1}{q+1} \left(1+\frac{1}{q+2}+\frac{1}{(q+2)(q+3)}+\cdots\right) \\ &<\frac{1}{q+1}\left(1+\frac{1}{q+1}+\frac{1}{(q+1)^{2}}+\cdots\right) \\ &=\frac{1}{q+1} \sum_{n=0}^{\infty} \frac{1}{(q+1)^{n}} \\ &=\frac{1}{q+1}\left(\frac{1}{1-\frac{1}{q+1}}\right) \\ &=\frac{1}{q}\end{aligned}\]
Since this is impossible, we conclude that no such integers \(p\) and \(q\) exist. \(\quad\) Q.E.D.
We call a real number which is not a rational number an irrational number.
We have seen that \(\sqrt{2}\) and \(e\) are irrational numbers.
Suppose \(\sum_{i=m}^{\infty} a_{i}\) and \(\sum_{i=k}^{\infty} b_{i}\) are infinite series for which there exists an integer \(N\) and a real number \(M>0\) such that \(0 \leq a_{i} \leq M b_{i}\) whenever \(i \geq N .\) If \(\sum_{i=k}^{\infty} b_{i}\) converges, then \(\sum_{i=m}^{\infty} a_{i}\) converges.
- Proof
-
Since \(\sum_{i=k}^{\infty} M b_{i}\) converges whenever \(\sum_{i=k}^{\infty} b_{i}\) does, the result follows from the comparison test. \(\quad\) Q.E.D.
Suppose \(\sum_{i=m}^{\infty} a_{i}\) diverges. Show that \(\sum_{i=m}^{\infty} \beta a_{i}\) diverges for any real number \(\beta \neq 0\).
Suppose \(\sum_{i=m}^{\infty} a_{i}\) and \(\sum_{i=k}^{\infty} b_{i}\) are infinite series for which there exists an integer \(N\) and a real number \(M>0\) such that \(0 \leq a_{i} \leq M b_{i}\) whenever \(i \geq N .\) If \(\sum_{i=m}^{\infty} a_{i}\) diverges, then \(\sum_{i=k}^{\infty} b_{i}\) diverges.
- Proof
-
By the comparison test, \(\sum_{i=m}^{\infty} M b_{i}\) diverges. Hence, by the previous exercise, \(\sum_{i=m}^{\infty} b_{i}\) also diverges.
We call the results of the next two exercises, which are direct consequences of the last two propositions, the limit comparison test.
Suppose \(\sum_{i=m}^{\infty} a_{i}\) and \(\sum_{i=m}^{\infty} b_{i}\) are infinite series for which \(a_{i} \geq 0\) and \(b_{i}>0\) for all \(i \geq m .\) Show that if \(\sum_{i=m}^{\infty} b_{i}\) converges and
\[\lim _{i \rightarrow \infty} \frac{a_{i}}{b_{i}}<+\infty ,\]
then \(\sum_{i=m}^{\infty} a_{i}\) converges.
Suppose \(\sum_{i=m}^{\infty} a_{i}\) and \(\sum_{i=m}^{\infty} b_{i}\) are infinite series for which \(a_{i} \geq 0\) and \(b_{i}>0\) for all \(i \geq m .\) Show that if \(\sum_{i=m}^{\infty} a_{i}\) diverges and
\[\lim _{i \rightarrow \infty} \frac{a_{i}}{b_{i}}<+\infty ,\]
then \(\sum_{i=m}^{\infty} b_{i}\) diverges.
Show that
\[\sum_{n=1}^{\infty} \frac{1}{n 2^{n}}\]
converges.
Show that
\[\sum_{n=0}^{\infty} \frac{x^{n}}{n !}\]
converges for any real number \(x \geq 0\).
Let \(S\) be the set of all finite sums of numbers in the set \(\left\{a_{1}, a_{2}, a_{3}, \dots\right\},\) where \(a_{i} \geq 0\) for \(i=1,2,3, \dots\) That is
\[S=\left\{\sum_{i \in J} a_{i}: J \subset\{1,2,3, \ldots, n\} \text { for some } n \in \mathbb{Z}^{+}\right\}.\]
Show that \(\sum_{i=1}^{\infty} a_{i}\) converges if and only if \(\sup S<\infty,\) in which case
\[\sum_{i=1}^{\infty} a_{i}=\sup S.\]
One consequence of the preceding exercise is that the sum of a sequence of nonnegative numbers depends only on the numbers begin added, and not on the order in which they are added. That is, if \(\varphi: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}\) is one-to-one and onto, \(\sum_{i=1}^{\infty} a_{i}\) converges if and only if \(\sum_{i=1}^{\infty} a_{\varphi(i)}\) converges, and, in that case,
\[\sum_{i=1}^{\infty} a_{i}=\sum_{i=1}^{\infty} a_{\varphi(i)}.\]