2. Introduction to Series¶
Series play a central role in calculus and are the main reason why we introduce sequences in the section above.
2.1. Definition¶
Series are simply defined by the sum of a sequence.
Series
A series is the sum of a sequence of terms, written as follows:
where \(m \le n\), and usually \(m=1\).
We can replace the upper value \(n\) with \(\infty\) for an infinite series. For example, the series of the infinite sequence \((u_1,u_2,u_3,\dots)\) is written as \(\displaystyle\sum_{i=1}^\infty u_i\).
We say that \(\displaystyle\sum_{i=m}^n u_i\) for \(n < \infty\) is a partial sum because the series is partially added up to just the \(n\)-th term.
Python code for the series of the sequence with recurrence relation \(u_k = u_{k-1}+\frac{1}{2}\) with \(u_1=0\):
n = 10 # number of terms
seq = 0 # first value of sequence
ser = 0 # first value of series
count = 0 # counter
while count < n: # while loop
print(ser) # print series value
seq = seq + 0.5 # update sequence
ser = ser + seq # update series
count += 1 # update counter
2.1.1. Summations¶
Character \(\Sigma\) is the Greek letter sigma, written in upper case, and denotes a summation.
We would read the expression above as “the sum of \(u_i\) from \(i=1\) to \(i=n\)”.
For example
is the sum of \(i\) from \(i=1\) to \(i=4\)
is the sum of \(2 i\) from \(i=1\) to \(i=4\)
From this example it should be clear that, in general:
where \(a\) is a constant.
However, note that
For example, \(\displaystyle \sum_{i=1}^4 (1+i) = 2 + 3 + 4 + 5 = 14\).
Examples
Write the following expressions using as a sum (\(\Sigma\) notation) without evaluating the sum:
\(1 + 5 + 9 + 13 + 17 +21\)
\(64-32+16-8+4-2+1\)
\(\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \dots + \frac{1}{99} \)
Solutions
\(\displaystyle\sum_{i=1}^6 (4n - 3)\) or \(\sum_{i=0}^5 (4n + 1)\).
\(\displaystyle\sum_{i=0}^6 (-1)^n 2^{6 - n}\) or \(\sum_{i=0}^6 (-1)^n 2^n\), by reversing the order of the summation.
\(\displaystyle\sum_{i=3}^{99} \frac{1}{n}\) or \(\sum_{i=1}^{97} \frac{1}{n+2}\). Note: other answers are possible!
2.2. Arithmetic and Geometric Progressions¶
We now introduce the arithmetic progression (otherwise known as the arithmetic sequence) and the geometric progression (otherwise known as the geometric sequence). These sequences have the nice property that their series can be solved.
Link to a short video on arithmetic and geometric progressions.
2.2.1. Arithmetic progression¶
2.2.1.1. Definition¶
Arithmetic Progression
An arithmetic progression (otherwise known as an arithmetic sequence) is a sequence of \(n\) terms which all have a “common difference” \(d\), written as follows:
where \(a\) is an arbitrary number. An example is the sequence \(3,5,7,9,11,13\) in which the first term is \(a=3\), the common difference is \(d=2\) and the number of terms is \(n=6\). An infinite arithmetic progression is given by the case where \(n \to \infty\).
The \(i\)-th term of the arithmetic progression can be expressed as the following recurrence relation:
for \(i=1,2,\dots,n\).
Python code for arithmetic progression:
n = 10 # number of terms
a = 30 # first value
d = -4 # commmon difference
seq = a # first value of sequence
count = 0 # counter
while count < n: # while loop
print(seq) # print series value
seq = seq + d # update sequence
count += 1 # update counter
You can play around with the behaviouur of different arithmetic progessions by changing the values of \(a\) and \(d\).
2.2.1.2. Solving the series of arithmetic progessions¶
Series of arithmetic progessions can be solved (i.e. reduced to a number). The series (or the sum) \(S_n\) of an arithmetic progression with recurrence relation \(u_i=a+(i-1)d\) is:
where \(L = a + (n-1) d\) is the last term in the series.
Notice in equation (2.5) that as \(n\) (the number of terms) increases, the terms of arithmetic progessions grow without bound. That is, as \(n\) approaches \(\infty\), the arithmetic progession approaches \(\infty\) if \(d>0\) and approaches \(-\infty\) if \(d<0\). In these cases we say that the series diverges (more on this later!). Infinite arithmetic progressions always diverge, the proof of which can be found via this link (although further reading is required!).
Python code for the series of the arithmetic progression:
nterms = 10 # number of terms
a = 30 # first value
d = -4 # commmon difference
seq = a # first value of sequence
ser = a # first value of series
count = 0 # counter
while count < nterms: # while loop
print(ser) # print series value
seq = seq + d # update sequence
ser = ser + seq # update series
count += 1 # update counter
You can use this code to convince yourself that the series of arithmetic progressions grow without bound for large \(n\).
Proof of the solution of the series of arithmetic progressions
From equation (2.5) write down the reversed series below the original:
Adding these two expressions gives:
Since there are \(n\) terms, this equations simplifies to the final result:
There is a legend, that Gauss (a famous mathematician) used a similar approach as a schoolboy back in the 1780s to find the answer to \(1+2+\dots+99+100\).
Examples
How many terms are in the sequence \(54, 52, 50,\dots, 18\)?
Find the sum of the first 30 terms of the arithmetic progression with the first three terms \(3,9,15\).
Find the sum of the arithmetic progression with first term 2, last term 10 and common difference 2.
Evaluate \(\sum_{n=1}^{10} (5n-1)\) and \(\sum_{n=100}^{1000} n\) .
In an arithmetic progression the 3rd term is 26 and the 8th term is 46. Find the 1st term, the common difference and \(S_{50}\).
Find the sum of the integers between 1 and 100 which are divisible by 6.
Solution
This is an arithmetic progression (AP) with first term \(a=54\), common difference \(d=-2\), and last term \(L=18\). Since \(L = a + (n-1)d\), we have \(n = \frac{18-54}{-2} = 19\) terms.
For an AP with first term \(a=3\), and common difference \(d=6\), the sum of the first \(30\) terms is given by \(S_{30} = \frac{n}{2}(2a+(n-1)d = \frac{30}{2} (6 + 29 \times 6) = 2700\).
For this AP the first term is \(a=2\), the common difference is \(d=2\). The last term is given by \(a+(n-1)d=10\), so we find that \(n=5\). The sum is given by \(S_5 = \frac{5}{2} (2+10) = 30\).
The first expression is an AP with first term \(a=4\), last term \(L=49\) and \(n=10\) terms. The sum is \(\frac{n}{2}(a+L)=265\). The second expression is an AP with first term \(n=100\), last term \(L=1000\) and \(n=901\) (take care!). The sum is \(\frac{n}{2} (a+L) = 495550\). You could also calculate this using \(\frac{1000}{2}(1+1000)−\frac{99}{2}(1+99)\).
Let the first term of this AP be denoted by \(a\), and the common difference be denoted by \(d\). We have that \(a+2d=26\) and \(a+7d=46\). Solving these two equations simultaneously gives \(a=18\) and \(d=4\). The sum of the first \(50\) terms is given by \(S_{50}=\frac{50}{2}(2a+49d)=5800\).
The series is given by \(S=6+12+18+...+96 = \sum_{n=1}^{16} 6 n\). This is the sum of an AP with first term \(a=6\), last term \(L=96\) and \(n=16\) terms. The result is \(S= \frac{16}{2}(6+96)=816\).
2.2.2. Geometric Progression¶
2.2.2.1. Definition¶
Geometric Progression
A geometric progression (otherwise known as an geometric sequence) is a sequence of \(n\) terms which have “common ratio” \(r\), written as follows:
where \(a\) is an arbitrary number. Using the notation for sequences in the previous section, the \(i\)-th term of the geometric progression is \(u_i=a r^{i-1}\) for \(i=1,2,\dots,n\). In other words, the next term in the geometric progression can be found by multiplying the previous term by the common ratio \(r\), given by the recurrence relation \(u_i = r u_{i-1}\) where \(u_1=a\). An example is the sequence \(24,−12,6,−3,1.5\) in which the first term is \(a=2\), the common ratio is \(r=−\frac{1}{2}\) and the number of terms is \(n=5\).
Python code for the geometric progression:
nterms = 10 # number of terms
a = 24 # first value
r = -0.5 # commmon difference
seq = a # first value of sequence
count = 0 # counter
while count < nterms: # while loop
print(seq) # print series value
seq = seq * r # update sequence
count += 1 # update counter
2.2.3. Solving the series of geometric progessions¶
Like arithmetic progessions, the series of geometric progressions can be solved (i.e. reduced to a number). The series (or the sum) \(S_n\) of a geometric progression \(u_i=a r^{i-1}\) is:
Python code for the series of the geometric progression:
nterms = 10 # number of terms
a = 24 # first value
r = -0.5 # commmon difference
seq = a # first value of sequence
ser = seq # first value of series
count = 0 # counter
while count < nterms: # while loop
print(seq) # print series value
seq = seq * r # update sequence
ser = ser + seq # update sequence
count += 1 # update counter
Proof of the solution of the series of geometric progressions
Starting from equation (2.6), the definition of the geometric progression:
we multiply by \(r\):
Subtracting \(r S_n\) from \(S_n\) and cancelling the terms provides:
which can be rearranged to provide the result.
2.2.4. A preliminary introduction to convergence and divergence¶
The behaviour of a geometric progression for “large” numbers of terms \(n\) depends on the value of the common ratio \(r\). This can be deduced by the following reasoning.
If \(r > 1\) or \(r < -1\) (which is to say that the absolute value \(\lvert r \rvert < 1\)) then the \(i\)-th term of the geometric progression \(u_i=a r^{i-1}\) becomes a very large positive or negative number. In this case we say that the series diverges.
Conversely, if \(\lvert r \rvert < 1\) then \(u_i=a r^{i-1}\) becomes a very small positive or negative number. In this case we say that the series converges.
The topics of divergence and convergence are explored in detail later in the course.
Thus we conclude that:
if \(\lvert r \rvert < 1\) then \(S_n\) converges and approaches \(\frac{a}{1-r}\) as \(n\to \infty\),
if \(\lvert r \rvert > 1\) then \(S_n\) diverges and approaches \(\infty\) as \(n\to \infty\).
Examples
Find the sum of the first five terms in a geometric progression with first term 27 and common ration is \(\frac{2}{3}\)?
Evaluate \( \sum_{n=1}^\infty 729 \Big(\frac{1}{3} \Big)^{n-1}\).
The first term of a geometric progression is \(8\). Given that the sum of the first three terms is 38, find the two possible values of the common ratio.
A geometric progression has first term \(27\) and common ratio of \(r=\frac{4}{3}\). Find the least number of terms that the sequence can have if its sum exceeds \(550\).
By writing the recurring decimal \(0.123123123\dots\) as a geometric progression, show that it can be expressed as the fraction \(\frac{41}{333}\).
Solutions
\(a=27\), \(r=\frac{2}{3}\), \(S_5 = a \frac{1 - r^5}{1-r} = 27 \frac{1 - (2/3)^5}{1 - 2/3} = \frac{211}{3}\),
The given expression represents the sum of a geometric progression (GP) with first term \(a=729\) and common ratio \(r=\frac{1}{3}\). The sum over all integers n is given by \(S_{\infty}=\frac{729}{1−1/3}=\frac{2187}{2}\).
Let the first term of this GP be denoted by \(a\) (\(=8\)), and the common ratio be denoted by \(r\). We have the sum of the first 3 terms is \(8(1+r+r^2)=38\). This equation is a quadratic with solutions \(r=−\frac{5}{2},\frac{3}{2}\).
The sum of the first \(n\) terms in this GP is given by \(S_n= 27 \frac{1−(4/3)^n}{1−4/3}\). We require that \(S_n>550\). The problem could be tackled by trial and error, but we will solve it analytically. The problem rearranges to give \((\frac{4}{3})^n>\frac{631}{81}\) (take care with the direction of the inequality!). By taking the natural logarithm we obtain \(n > \frac{\ln \big( \frac{631}{81} \big)}{ \ln \big( \frac{4}{3} \big)}=7.136\) (4sf). Since \(n\) is an integer, the least number of terms required is \(8\).
2.3. Method of Differences¶
The method of differences provides a way to find a finite series (sum of a sequence) by using the difference of similar sums to compute the desired result. This is a handy trick to find the value of partial (finite) sums.
This is best illustrated by example.
2.3.1. Illustrative example¶
Say we want to compute the partial sum:
One way to approach this is to consider the following two series:
It can be seen that in general the \(n\)-th term in the second series is the same as the \((n-1)\)-th term in the first series. Therefore, if we calculate the difference between the two sums, almost all of the terms cancel, to leave:
Observing that \(r (r+1) - r (r-1) = 2 r\) then gives us the following result:
We note that this result could also have been found using the formula for an arithmetic progression with first term 1 and common difference 1.
2.3.2. Rewriting the summation index¶
2.3.2.1. Demonstration of utility¶
It is possible to demonstrate the cancellation of terms without writing out the terms in the series. Using the example above, we could rewrite:
and therefore:
2.3.2.2. Generalisation¶
In general, rewriting the summation index provides:
where \(f(r)\) is an arbitrary function of summation index \(r\) and \(c\) is a postive integer number. This can be also used for negative integer values:
You will find that this technique can drastically simplify summations we wish to solve. For example:
2.3.2.3. Example¶
Consider the difference between the following two sums:
Notice that the function within summation on the right \(\frac{e^{r+2}}{(r+3) (r+4)}\) is the same as that on the left \(\frac{e^r}{ (r+1) (r+2) }\) except that the summation index is greater by a value of 2. Thus we can rewrite the summation index of the right sum to provide:
Now we have the same function of the index \(r\) but with different values of the summation index. To solve the equation we simply “take out” the index values that are not common, i.e. \(r=1,2\) on the left sumation and \(r=N+1,N+2\) on the right summation:
Note: if you struggle with rewriting the summation index, it’s ok to just write out the terms in the series and demonstrate the cross-cancellation.
Questions
Use the result \(r^2 (r+1)^2 - (r-1)^2 r^2 = 4 r^3\) to calculate \(\sum_{r = 1}^{N} r^3\).
By using partial fractions to rewrite the summand, show the result:
Solutions
\(4 \sum_{r=1}^N r^3 = \sum_{r=1}^N \big( r^2(r+1)^2 - (r-1)^2 r^2 \big)\) \(= \sum_{r=1}^N r^2(r+1)^2 - \sum{r=1}^N (r-1)^2 r^2 \) \(= 1^2 2^2 + 2^2 3^2 + \dots + (N-1)^2 N^2 + N^2 (N+1)^2\) \( - \big( 0^2 1^2 + + 1^2 2^2 + 2^2 3^2 + \dots + (N-1)^2 N^2 \big)\). Since all terms cancel except for \(N^2 (N+1)^2\) we have that \(\sum_{r=1}^N r^3 = \frac{N^2 (N+1)^2}{4}\).
Let \( \frac{2}{2(r+1)(r+2)} = \frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+2}\) \(= \frac{A(r+1)(r+2) + B r (r+2) + Cr (r+1)}{ r (r+1) (r+2)}\). By equating coefficients of \(r^0\), \(r^1\) and \(r^2\) in the numerator, we obtain: \(A=1\), \(B=-2\), \(C=1\). Then \(\sum_{r=1}^n \frac{2}{2(r+1)(r+2)}\) \(=\sum_{r=1}^n \frac{1}{r} - 2 \sum_{r=1}^n \frac{1}{r+1} + \sum_{r=1}^n \frac{1}{r+2}\) \( = \sum_{r=1}^n \frac{1}{2} - 2 \frac{r=2}{n+1} \frac{1}{r} + \sum_{r=3}^{n+2} \frac{1}{r}\) \( = (1-2+1) \sum_{r=3}^n \frac{1}{r} + \big( \frac{1}{1} + \frac{1}{2} \big) -2 \big( \frac{1}{2} + \frac{1}{n+1} \big) +\big( \frac{1}{n+1} + \frac{1}{n+1} \big)\) \(=\frac{1}{n+2} - \frac{1}{n+1} + \frac{1}{2}\).