Example 5.28.
The series \(\sum 1 =1 +1 + \cdots \) diverges to \(+\infty\) since its sequence of partial sums is the sequence \(1,2,3,\ldots\)
Section 5.3 Series
A series is the sum of a sequence. It is clear what that means if the sequence is finite. We use the symbol \(\sum_{n=m}^k a_n\) as a shorthand for the sum
\begin{equation*}
a_m + a_{m+1} + \cdots + a_k.
\end{equation*}
For an infinite sequence \((a_n)\text{,}\) we need to make sense of what means by its sum \(\sum_{n=1}^\infty a_n\text{.}\) To do so, we consider the sequence of partial sums of the series \(\sum a_n\text{:}\) for \(k \ge 1\text{,}\) let \(s_k\) be the sum of the first \(k\) terms of \((a_n)\text{.}\) That is,
\begin{equation*}
s_1 = a_1, s_2=a_1+a_2, \ldots, s_k = \sum_{n=1}^k a_n, \ldots
\end{equation*}
We say that a series \(\sum a_n\) is convergent if \((s_k)\) its sequence of partial sums does. In that case, we write \(\sum a_n = L\) and say that the series \(\sum a_n\) sums to \(L\) where \(L\) is the limit of \((s_k)\text{.}\) Likewise, a series \(\sum a_n\) diverges (resp. diverges (to \(\pm\infty\)) if its sequence of partial sums does so.
Example 5.29.
The series \(\sum (-1)^n\) is divergent because \(s_k\text{,}\) its \(k\)th partial sum is \(-1\) when \(k\) is odd and is \(0\) when \(k\) is even and we have seen that the sequence \((s_k)\) is divergent.
The divergence of the series in the examples above can also be deduced from the following result.
Proposition 5.30. Divergence Test.
Proof.
Suppose \((s_k)\) its sequence of partial sums converges to some \(L\text{.}\) Then for any \(n\text{,}\)
\begin{equation*}
0 \le |a_n| = |s_{n} -s_{n-1}| \le |s_n -L| + |s_{n-1}-L|
\end{equation*}
Since both \(|s_n-L|\) and \(|s_{n-1}-L|\) converges to 0, therefore so is \((a_n)\) by the Squeeze Lemma.
Since convergence of series is the converges of its sequence of partial sums, some algebraic properties of convergent series follows directly from their sequence counterparts. For example,
Proposition 5.31.
If \(\sum a_n = A\) and \(\sum b_n =B\text{,}\) then-
\(\sum (a_n+b_n) = A + B\text{.}\)
-
\(\displaystyle \sum ca_n = cA.\)
It is not true that \(\sum a_nb_n = AB\text{.}\) For instance, let \(a_n =
1/2^{n} =b_n\text{,}\) then \(\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty}
b_n = 1 \text{.}\) But, \(\sum_{n=1}^\infty a_nb_n = \sum_{n=1}^\infty
(1/4)^n = 1/3\)
Also, the proposition can be extended to the cases including \(A,B = \pm
\infty\) if some pretty self-evident rules are followed: \((+\infty)+(+\infty) = +\infty\text{,}\) \((-\infty)+(-\infty) = -\infty\text{.}\) \(c(\infty) = +\infty\) if \(c \gt 0\text{,}\) \(= -\infty\) if \(c \lt 0\text{.}\) If \(\sum a_n\) diverges to \(+\infty\) and \(\sum b_n\) diverges to \(-\infty\text{,}\) then the convergence of \(\sum (a_n+b_n)\) needs further investigation.
A series \(\sum a_n \) is telescopic if \(a_n = b_n -
b_{n+m}\) for some sequence \((b_n) \) and some \(m \ge 1\text{.}\) For \(k \ge m\text{,}\) the \(k\)-th partial sum of a telescopic series \(\sum a_n\) looks like
\begin{align*}
s_k & = \sum_{n=1}^k a_n = \sum_{n=1}^k b_n - b_{n+m} \\
& =\sum_{n=1}^k b_n - \sum_{n=1}^k b_{n+m} =\sum_{n=1}^k b_n
- \sum_{n=m+1}^{k+m} b_{n}\\
& =(b_1 + \cdots + b_m) -
(b_{k+1}+ \cdots + b_{k+m})
\end{align*}
So, \(\sum a_n\) converges if and only if \((b_{k+1}+ \cdots +
b_{k+m})\) converges as \(k \to \infty\text{.}\) So, if \((b_n) \to B\text{,}\) then \(\sum a_n\) converges to \(b_1+\cdots + b_m - mB\text{.}\)
Example 5.32.
Let us find the sum of the series
\begin{equation*}
\ds \sum \left(\frac{3n+1}{n} - \frac{3n+7}{n+2}\right)\text{.}
\end{equation*}
This is a telescopic series with its \(n\)-th term equals \(b_n -b_{n+2}\) where \(b_n = (3n+1)/n\text{.}\) Since, \(b_n \to 3\text{,}\) it follows from the discuss above that the series converges to \(b_1 + b_2 - 2(3) = 4 + 7/2 - 6 = 3/2\text{.}\)
Example 5.33.
The sum of the reciprocal of the triangular numbers is telescopic
\begin{equation*}
\sum_{n \ge 1} \frac{1}{T_n} = \sum_{n \ge 1} \frac{2}{n(n+1)} =
2 \sum_{n \ge 1} \left(\frac{1}{n} - \frac{1}{n+1}\right).
\end{equation*}
And its \(k\)-partial sum is \(2(1-1/(k+1))\text{.}\) Hence \(\sum 1/T_n\) converges to \(2\text{.}\)
Geometric Series are sums of geometric sequences 5.6. They form an important class of series whose sequence of partial sums can be computed readily.
Proposition 5.34.
Suppose \(a \neq 0\text{.}\) Then
\begin{equation*}
\sum_{n=0}^{\infty} ar^n =
\begin{cases}
\dfrac{a}{1-r} & |r| \lt 1;\\
\text{is divergent} & |r| \ge 1
\end{cases}
\end{equation*}
Proof.
Since \(a \neq 0\text{,}\) if \(r=1\text{,}\) then \(\sum_{n=0}^{\infty}
ar^n =\sum_{n=0}^{\infty} a \) is divergent by Proposition 5.30.
So, let us assume \(r \neq 1\text{.}\) Let \(s_k\) be the \(k\)-th partial sum of the series. So, \(s_k = \sum_{n=0}^{k-1} ar^n\) and we have
\begin{equation*}
s_k -rs_k = \sum_{n=0}^{k=1} ar^n - \sum_{n=0}^{k-1}
ar^{n+1} = a-ar^{k} = a(1-r^k).
\end{equation*}
Therefore, \(s_k = \dfrac{a(1-r^{k})}{1-r}\text{.}\) If \(|r| \lt 1\text{,}\) then \(r^{k} \to 0\) (Proposition 5.18) and so the series sums to \(\dfrac{a}{1-r}\text{.}\) If \(|r| \gt 1\text{,}\) the sequence \((r^{k})\) (Proposition 5.18) is unbounded, and so is \((s_k)\text{.}\) Therefore, the series diverges.
Example 5.35.
Taking \(a = 1/2= r\text{,}\) we get from the proposition above
\begin{equation*}
\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots =
\sum_{n=1}^{\infty}\frac{1}{2^k} =
\frac{\frac{1}{2}}{1-\frac{1}{2}} = 1.
\end{equation*}
Read
this page for the relation of this series with a Zeno paradox.
