Numerical series

Part 5 · Series · Chapter 1 · lecture notes by Fabio Furini · Chapter PDF

1. Numerical series

• We now introduce numerical series, which extend the operation of addition to an infinite number of terms.

The sum of infinitely many terms, even if they are all positive, can give a finite result.

• Imagine measuring a square of area 2 using the following procedure: we divide the square in half along the diagonal and measure the first right triangle: we obtain 1; then we divide the second right triangle in half and measure the first of the two resulting right triangles: we obtain \(\frac{1}{2}\); the remaining right triangle is again divided in half... and so on indefinitely. We obtain the infinite sum:

  \[ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + {\rm \dots} + \frac{1}{2^k} + {\rm \dots} = \sum_{k=0}^{\infty} \frac{1}{2^k} \]

  which, by the way it was constructed, must give 2 as its result.

Definition 1: of numerical series

Given a sequence \(\{a_k\}_{k \in \N}\), we call numerical series of the terms \(a_k\) the expression:

\[ \sum_{k=0}^{\infty} a_k \]

• It is read “series (or also sum) for \(k\) from 0 to \(\ip\) of \(a_k\)”. The values \(a_k\) are called the general terms of the series.

Definition 2: of sequence of partial sums

The numbers

\[ s_n = \sum_{k=0}^{n} a_k = a_0 + a_1 + {\rm \dots} + a_n,~~~~ \forall n \in \N \]

are called the \(n\)-th partial sums of the series and they define the sequence \(\{s_n\}\) of partial sums.

The behavior (character) of the series is determined by the limit of the sequence \(\{ s_n\}\) as \(n\) tends to infinity. We say that a series is convergent, divergent, irregular, if the sequence \(\{s_n\}\) of partial sums is convergent, divergent or irregular, respectively.

Definition 3: of sum of the series

If the sequence \(\{s_n\}\) of partial sums is convergent, that is, if

\[ s_n \rr s \in \R {\rm ~~~~as~~~~}n \rr \ip \]

we say that \(s\) is the sum of the series, and we write \(\sum_{k=0}^{\infty} a_k = s\).

Hence, if the sequence \(\{s_n\}\) is convergent, we have:

\[ \sum_{k=0}^{\infty} a_k = \lim_{n \rr \ip} \sum_{k=0}^{n} a_k = \lim_{n \rr \ip} s_n = s \]

The series makes precise the idea of a sum of infinitely many terms, that is, we compute the limit as \(n \rr \ip\) of the finite sum of the first \(n\) terms.

• If, instead of summing starting from \(0\), we start from an index \(n_0 >0\), we write \(\sum_{k=n_0}^{\infty} a_k\)

• To denote a generic numerical series without specifying the starting index \(n_0\) we will use the symbol \(\sum a_k\)

A generic numerical series \(\sum a_k\) always involves two different sequences:

1. the sequence \(\{a_k\}\) of the terms of the series

2. the sequence \(\{ s_n\}\) of partial sums

Example 1: behavior of a series

Let us determine the behavior of the series:

\[ \sum_{k=1}^{\infty} \frac{1}{\sqrt{k}} \]

We have:

\[ s_n = \sum_{k=1}^{n} \frac{1}{\sqrt{k}} = \underbrace{1 + \frac{1}{\sqrt{2}}+{\dots}+\frac{1}{\sqrt{n}}}_{n {\rm ~terms,~each~} \ge \frac{1}{\sqrt{n}} } \ge n \cdot \frac{1}{\sqrt{n}} = \sqrt{n} \rr \ip {\rm ~~~as~~~} n \rr \ip \]

Hence the sequence of partial sums \(\{s_n\}\) is divergent by the comparison theorem for sequences and, consequently, the series is divergent.

1.1 Main properties of numerical series

Remark 1

If a sequence \(\{a_k\}\) has non-negative terms, that is, \(a_k \ge 0,\forall k\), then the sequence of partial sums \(\{s_n\}\) is increasing and regular.

Proof

Given a sequence \(\{a_k\}\) with non-negative terms, the sequence of partial sums \(\{s_n\}\) is increasing since:

\[ s_{n+1}= s_{n} + \underbrace{a_{n+1}}_{\ge 0} \ge s_{n}, ~~~\forall n, {\rm ~~~~hence~~} \lim_{n \rr \ip} s_n = \sup_{n \in \N} \{s_n\} \]

by the monotonicity theorem for sequences. Consequently the sequence \(\{s_n\}\) cannot be irregular. It is therefore regular, that is, it either converges or diverges. □

Given a sequence \(\{a_k\}\) with non-negative terms:

• If \(\{s_n\}\) is increasing and bounded, then it has a finite limit and \(\sum a_k\) converges.

• If \(\{s_n\}\) is increasing and unbounded, then \(\sum a_k\) diverges to \(\ip\).

• This remark also holds for sequences with eventually non-negative terms, positive terms, or eventually positive terms.

Theorem 1

If a series \(\sum a_k\) is convergent, then \(\lim_{k \rr \ip} a_k=0\)

Proof

By definition of convergent series, the sequence of partial sums \(\{s_n\}\) converges to a real number \(s\), i.e., \(s_n \rr s \in \R\) as \(n \rr \ip\). Without loss of generality we consider \(n_0=0\).

We observe that the sequence \(\{s_n\}\) can be defined recursively:

\[ s_0= a_0,~~~~ s_n = s_{n-1} + a_n,~~ \forall n \ge 1 \]

Consequently we have:

\[ a_n = s_n - s_{n-1} {\rm ~~~~and~hence~~~} \lim_{n \rr \ip} a_n = \lim_{n \rr \ip} \big( s_n -s_{n-1} \big)= s -s =0 \]

□

The convergence of a series therefore implies that \(a_k \rr 0\) as \(k \rr \ip\), that is:

\[\begin{equation} \sum a_k {\rm ~~convergent~~} ~~\Rightarrow~~ \lim_{k \rr \ip} a_k=0 \label{BBB} \end{equation}\]

but the converse is not true (Example 1 is a counterexample):

\[\begin{equation} \lim_{k \rr \ip} a_k=0 ~~\nRightarrow~~ \sum a_k {\rm ~~convergent~~} \label{CCC} \end{equation}\]

Hence the fact that \(a_k \rr 0\) as \(k \rr \ip\) is a necessary but not sufficient condition for the convergence of a series.

• From the contrapositive of \(\eqref{BBB}\) we have:

  \[ \lim_{k \rr \ip} a_k \neq 0 ~~\Rightarrow~~ \sum a_k {\rm ~is~divergent~or ~irregular~~} \]

  that is, if the limit is not equal to zero, then the series is not convergent.

1.2 Tails of series

If we modify a finite number of terms of a series, the value of the sum may change, but the behavior of the series remains unchanged.

• If we modify the value of one term, for example \(a_{n_0}\), for \(n \ge n_0\) the sequences of partial sums, the original and the modified one, differ only by that term: hence both converge, both diverge, or both are irregular. The same holds if we modify a finite number of terms.

The results on the behavior of series therefore also hold if the hypotheses are satisfied “eventually”, i.e., from a certain index \(n_0\) onward.

Definition 4: of tail of a series

Given a series \(\sum_{k=0}^{\infty} a_k\) and a value \(m \in \N\), the series \(\sum_{k=m+1}^{\infty} a_k\) is called a tail of the series.

• Given a series \(\sum_{k=0}^{\infty} a_k\) and \(m \in \N\) we have:

  \[\begin{equation} \sum_{k=m+1}^{n} a_k = \left(\sum_{k=0}^{n} a_k\right) - \left(\sum_{k=0}^{m} a_k\right) = s_n -s_m, \qquad \forall n > m \label{MMMMM} \end{equation}\]

  hence for every tail (that is, for every \(m\)) the associated sequence of partial sums is equal to that of the original series minus a constant. Consequently we have the following remark:

Every tail of a series has the same behavior as the original series.

• Given a convergent series, taking the limit as \(n \rr \ip\) in \(\eqref{MMMMM}\) we obtain the relation:

  \[\begin{equation} \sum_{k=m+1}^{\infty} a_k = \left(\sum_{k=0}^{\infty} a_k\right) - \left(\sum_{k=0}^{m} a_k\right) = s -s_m~~~ {\rm ~~as~~} n \rr \ip \label{NNNNN} \end{equation}\]

  Consequently we have the following remark:

Given \(m \in \N\), every tail \(\sum_{k=m+1}^{\infty} a_k\) of a convergent series can be interpreted as the error that we make when approximating the sum \(s\) with the partial sum \(s_m\).

Theorem 2

If a series \(\sum a_k\) is convergent, then

\[ \underbrace{s-s_m}_{\displaystyle=\sum_{k=m+1}^{\infty} a_k} \rr 0 {\rm ~~~as~~~} m \rr \ip \]

Proof

For every \(n > m\), the \(n\)-th partial sum of the series is:

\[ s_n = s_{m} + \sum_{k=m+1}^{n} a_k \]

Taking the limit as \(n \rr \ip\), we have:

\[ s = s_{m} + \underbrace{\lim_{n \rr \ip}\sum_{k=m+1}^{n} a_k }_{\displaystyle =\sum_{k=m+1}^{\infty} a_k} \]

Now taking the limit as \(m \rr \ip\), we have:

\[ s = s + \lim_{m \rr \ip} \sum_{k=m+1}^{\infty} a_k ~~~\Rightarrow~~~ \lim_{m \rr \ip} \sum_{k=m+1}^{\infty} a_k = 0 \]

that is, the claim of the theorem. □

This theorem tells us that, if a series is convergent, then the value \(s - s_m\), that is, the error that we make when approximating the sum \(s\) with the partial sum \(s_m\), tends to zero as \(m \rr \ip\). In other words, the tail of a convergent series tends to zero as \(m\) tends to infinity.

1.3 Harmonic series

Definition 5: of harmonic series

The harmonic series is the series \(\sum_{k=1}^{\infty} \frac{1}{k}\)

Theorem 3: behavior of the harmonic series

The harmonic series diverges to \(\ip\)

• Graphically we have:

The harmonic series \(\sum_{k=1}^{\infty} \frac{1}{k}\) is a counterexample that proves \(\eqref{CCC}\), since:

\[ \lim_{k \rr \ip} \frac{1}{k} = 0 {\rm ~~~~~~~and~~~~~~~} \sum_{k=1}^{\infty} \frac{1}{k}{\rm ~~~~~diverges~to} \ip \]

• Before proving the theorem, we observe that for certain values of \(n\) the partial sums are:

  \[ \underbrace{s_1}_{\displaystyle =s_{2^{\red 0}}}=1+\frac{\red 0}{2},~~~~\underbrace{s_2}_{\displaystyle =s_{2^{\red 1}}}= s_1 +\frac{1}{2}=1+\frac{0}{2}+\frac{1}{2} = 1+\frac{\red 1}{2},~~~~\underbrace{s_4}_{\displaystyle =s_{2^{\red 2}}}=s_2 + \underbrace{\left(\frac{1}{3}+\frac{1}{4}\right)}_{\ge \frac{1}{4}+\frac{1}{4}=\frac{1}{2}}\ge 1+\frac{1}{2} + \frac{1}{2}=1 + \frac{\red 2}{2} \]
  \[ \underbrace{s_8}_{\displaystyle =s_{2^{\red 3}}}=s_4 + \underbrace{\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)}_{\ge \frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}} \ge 1 + \frac{2}{2} + \frac{1}{2} = 1 + \frac{\red 3}{2} \]
  \[ \underbrace{s_{16}}_{\displaystyle =s_{2^{\red 4}}}=s_8 + \underbrace{\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+ \frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16} \right)}_{\ge \frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{1}{2}} \ge 1 + \frac{3}{2} + \frac{1}{2} = 1 + \frac{\red 4}{2} \]

Proof

We prove by induction on \(2^n\) that

\[ s_{2^n} \ge 1 + \frac{n}{2},~~~ \forall n \in \N \]

Base case. Let \(n = 0\). Then the statement becomes \(s_{2^0} \ge 1+\frac{0}{2}\), i.e., \(1 \ge 1\), which is clearly true.

Inductive step. Assume that it is true for \(2^{n-1}\) and let us prove it for \(2^{n}\). By the inductive hypothesis, we have:

\[ s_{2^{n-1}} \ge 1 + \frac{n-1}{2} \]

Moreover, we have:

\[ s_{2^{n}} = s_{2^{n-1}} + \underbrace{\sum_{k=2^{n-1}+1}^{2^n} \left( \frac{1}{k} \right)}_{\displaystyle \ge 2^{n-1} \cdot \frac{1}{2^n}=\frac{1}{2}} {\rm ~~~~hence~~~~~} s_{2^{n}} \ge 1 + \frac{n-1}{2} + \frac{1}{2} = 1 +\frac{n}{2} \]

which is exactly the desired statement for \(2^{n}\).

We therefore have:

\[ s_{2^n} \ge 1 + \frac{n}{2}\rr \ip {\rm ~~~as~~~} n \rr \ip \]

and \(\{s_n\}\) is unbounded above and also monotonically increasing, since \(a_k=1/k>0, \forall k >1\). Consequently \(\{s_n\}\) diverges to \(\ip\) by the monotonicity theorem for sequences, and the harmonic series diverges to \(\ip\). □

The sequence \(a_k=1/k, \forall k \ge 1\), of the harmonic series has positive terms, hence the sequence of partial sums \(\{s_n\}\) is increasing, and it is divergent since it is unbounded.

1.4 Geometric series

Definition 6: of geometric series

Given \(q \in \R\), the geometric series with ratio \(q\) is the series \(\sum_{k=0}^{\infty} q^k\)

Theorem 4: behavior and sum of the geometric series

Given the ratio \(q \in \R\),

\[ {\rm the~geometric~series~} \sum_{k=0}^{\infty} q^k {\rm ~~~~is~~~~} \begin{cases} {\rm convergent~} & {\rm if~} |q| < 1\\[1ex] {\rm divergent~to~} \ip & {\rm if~} q \ge 1\\[1ex] {\rm irregular~} & {\rm if~} q \le -1 \end{cases} \]

If the geometric series is convergent, its sum \(s\) is \(\frac{1}{1-q}\)

Proof

Given \(q \in \R\) and \(n \in \N\), the \(n\)-th partial sum of the geometric series is:

\[ \sum_{k=0}^n q^k = \begin{cases} \displaystyle \frac{1-q^{n+1}}{1-q} & {\rm if~} q \neq 1\\[3ex] n+1 & {\rm if~} q = 1 \end{cases} \]

since it equals the sum of the first \(n+1\) terms of the geometric progression. Moreover, we have:

\[ \lim_{n \rightarrow +\infty} q^n = \begin{cases} +\infty & {\rm if~} q >1\\[2ex] 1 & {\rm if~} q = 1\\[2ex] 0 & {\rm if~} |q| < 1\\[2ex] {\rm does~not~exist~} & {\rm if~} q \le -1 \end{cases} \]

Hence, if \(q \neq 1\):

\[ \lim_{n \rightarrow +\infty} s_n = \lim_{n \rightarrow +\infty} \frac{1-q^{n+1}}{1-q} = \frac{1}{1-q} \; \lim_{n \rightarrow +\infty} \big(1-q^{n+1}\big) = \begin{cases} \displaystyle \frac{1}{1-q} & {\rm if~} |q| < 1\\[2ex] +\infty & {\rm if~} q > 1\\[2ex] {\rm does~not~exist~} & {\rm if~} q \le -1 \end{cases} \]

and if \(q=1\) we have:

\[ \lim_{n \rightarrow +\infty} s_n = \lim_{n \rightarrow +\infty} (n+1)= +\infty \]

□

Example 2: geometric series

Let us determine the behavior of the series:

\[ \sum_{k=0}^{\infty} \frac{1}{2^k} = \sum_{k=0}^{\infty} \left(\frac{1}{2}\right)^k \]

It is a geometric series with ratio \(q=1/2\), hence it is convergent and its sum \(s\) is \(1/(1-1/2) =2\).

Let us determine the behavior of the series:

\[ \sum_{k=0}^{\infty} \left(\frac{13}{12}\right)^k \]

It is a geometric series with ratio \(q=13/12\), hence it is divergent.

Try it — the interactive graph below shows what you have just read: move the sliders.

1.5 Telescoping series

Definition 7: of telescoping series

A telescoping series is a series of the form:

\[ \sum_{k=n_0}^{\infty} (b_k-b_{k+1}) {\rm ~~~~~where~~} \{b_k\} {\rm~is~a ~sequence} \]

Theorem 5: behavior and sum of the telescoping series

A telescoping series converges, diverges or is irregular according to whether the sequence \(\{b_k\}\) converges, diverges or is irregular, respectively.

\[ {\rm ~~If~~} \lim_{k \rr \ip} b_k = \ell \in \R^* {\rm ~~~~then~~~~} \sum_{k=n_0}^{\infty} (b_k-b_{k+1})= b_{n_0} - \underbrace{\lim_{k \rr \ip} b_k}_{=\ell \in \R^*} \]

Moreover, if \(\ell \in \R\), then the telescoping series is convergent and its sum \(s\) is \(b_{n_0} - \ell\)

Proof

We have

\[\begin{align*} s_n &= \sum_{k=n_0}^{n_0+n} \big(b_k - b_{k+1}\big)\\[2ex] & = (b_{n_0} - b_{n_0+1}) + (b_{n_0+1} - b_{n_0+2}) + {\rm \dots} + (b_{n_0+n} - b_{n_0+n+1})\\[2ex] &= b_{n_0} - b_{n_0+n+1} \end{align*}\]

hence:

\[ \sum_{k=n_0}^{\infty} (b_k-b_{k+1})=\lim_{n \rightarrow +\infty} s_n = \lim_{n \rightarrow +\infty} (b_{n_0} - b_{n_0+n+1}) = b_{n_0} -\lim_{n \rr \ip} b_n \]

□

Remark 2

The series \(\sum_{k=1}^{\infty} \frac{1}{k \; (k+1)}\), called Mengoli's series, is convergent and its sum \(s\) is equal to \(1\).

Proof

We have:

\[ \sum_{k=1}^{\infty} \frac{1}{k \; (k+1)} = \sum_{k=1}^{\infty} \frac{(k+1) - k }{k \; (k+1)} = \sum_{k=1}^{\infty} \underbrace{\frac{1}{k}}_{=b_k} - \underbrace{\frac{1}{k+1}}_{=b_{k+1}} \]

Mengoli's series therefore has the form:

\[ b_k - b_{k+1} {\rm ~~~with~~~} b_k = \frac{1}{k} {\rm ~~~~~and~moreover~~} \lim_{k \rr \ip} b_k = \lim_{k \rr \ip} \frac{1}{k} = 0 \]

hence it is a convergent telescoping series (\(n_0=1\)) and moreover \(s= b_1 = 1\). □

• Graphically we have:

Proof

Alternative proof

For every \(n\ge 1\), the value of the partial sum of Mengoli's series is:

\[ s_n = \sum_{k=1}^{n} \frac{1}{k \; (k+1)}= \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+1} \right)= \left(1 -\frac{1}{2} \right) + \left(\frac{1}{2} -\frac{1}{3} \right) + {\rm \dots} + \left(\frac{1}{n} -\frac{1}{n+1} \right)= 1 - \frac{1}{n+1} \]

Hence we have

\[ \sum_{k=1}^{\infty} a_k = \lim_{n \rr \ip} s_n = \lim_{n \rr \ip} 1 - \frac{1}{n+1} = 1 \]

Consequently Mengoli's series is convergent and its sum \(s\) is 1. □

Example 3: telescoping series

Let us determine the behavior of the series:

\[ \sum_{k=1}^{\infty} \frac{1}{(3\;k+2)\;(3\;k+5)} \]

We have:

\[ \frac{1}{(3\;k+2)\;(3\;k+5)} = \frac{\alpha}{3\;k+2}-\frac{\beta}{3\;k+5} = \frac{\alpha\:(3\;k+5)-\beta\;(3\;k+2)}{(3\;k+2)\;(3\;k+5)} = \frac{3\;(\alpha-\beta)\;k +5\;\alpha -2\;\beta}{(3\;k+2)\;(3\;k+5)} \]

\[ \Rightarrow \begin{cases} 3\;(\alpha-\beta) =0\\ 5\;\alpha -2\;\beta =1 \end{cases} \Rightarrow \alpha=\beta,~~~ 3\;\alpha=1 {\rm ~~hence~~} \alpha=\beta=\frac{1}{3} \]

consequently:

\[ \frac{1}{(3\;k+2)\;(3\;k+5)} = \frac{1}{3} \left(\frac{1}{3\;k+2} - \frac{1}{3\;k+5} \right) = \frac{1}{3} \left( \underbrace{\frac{1}{3\;k+2}}_{=b_k} - \underbrace{\frac{1}{3\;(k+1)+2}}_{=b_{k+1}} \right) \]

The series therefore has the form:

\[ b_k - b_{k+1} {\rm ~~~with~~~} b_k = \frac{1}{3\;k+2} {\rm ~~~~~and~moreover~~} \lim_{k \rr \ip} b_k = \lim_{k \rr \ip} \frac{1}{3\;k+2} = 0 \]

that is, a telescoping series (\(n_0=1\)), and we have:

\[\begin{align*} \sum_{k=1}^{\infty} \frac{1}{(3\;k+2)\;(3\;k+5)} = \frac{1}{3} \left( b_1 - \underbrace{ \lim_{k \rr \ip} b_{k}}_{\rr 0}\right) = \frac{1}{3} \; b_1 = \frac{1}{15} \end{align*}\]

Hence it is a convergent telescoping series and its sum \(s\) is \(\frac{1}{15}\).