# Summation of series

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Summation of series
```SERIES AND LIMITS
some sort of relationship between successive terms. For example, if the nth term
of a series is given by
un =
1
,
2n
for n = 1, 2, 3, . . . , N then the sum of the ﬁrst N terms will be
SN =
N
n=1
un =
1
1 1 1
+ + + ··· + N.
2 4 8
2
(4.1)
It is clear that the sum of a ﬁnite number of terms is always ﬁnite, provided
that each term is itself ﬁnite. It is often of practical interest, however, to consider
the sum of a series with an inﬁnite number of ﬁnite terms. The sum of an
inﬁnite number of terms is best deﬁned by ﬁrst considering the partial sum
of the ﬁrst N terms, SN . If the value of the partial sum SN tends to a ﬁnite
limit, S, as N tends to inﬁnity, then the series is said to converge and its sum
is given by the limit S. In other words, the sum of an inﬁnite series is given
by
S = lim SN ,
N→∞
provided the limit exists. For complex inﬁnite series, if SN approaches a limit
S = X + iY as N → ∞, this means that XN → X and YN → Y separately, i.e.
the real and imaginary parts of the series are each convergent series with sums
X and Y respectively.
However, not all inﬁnite series have ﬁnite sums. As N → ∞, the value of the
partial sum SN may diverge: it may approach +∞ or −∞, or oscillate ﬁnitely
or inﬁnitely. Moreover, for a series where each term depends on some variable,
its convergence can depend on the value assumed by the variable. Whether an
inﬁnite series converges, diverges or oscillates has important implications when
describing physical systems. Methods for determining whether a series converges
are discussed in section 4.3.
4.2 Summation of series
It is often necessary to ﬁnd the sum of a ﬁnite series or a convergent inﬁnite
series. We now describe arithmetic, geometric and arithmetico-geometric series,
which are particularly common and for which the sums are easily found. Other
methods that can sometimes be used to sum more complicated series are discussed
below.
116
4.2 SUMMATION OF SERIES
4.2.1 Arithmetic series
An arithmetic series has the characteristic that the diﬀerence between successive
terms is constant. The sum of a general arithmetic series is written
SN = a + (a + d) + (a + 2d) + · · · + [a + (N − 1)d] =
N−1
(a + nd).
n=0
Rewriting the series in the opposite order and adding this term by term to the
original expression for SN , we ﬁnd
SN =
N
N
[a + a + (N − 1)d] = (ﬁrst term + last term).
2
2
(4.2)
If an inﬁnite number of such terms are added the series will increase (or decrease)
indeﬁnitely; that is to say, it diverges.
Sum the integers between 1 and 1000 inclusive.
This is an arithmetic series with a = 1, d = 1 and N = 1000. Therefore, using (4.2) we ﬁnd
1000
(1 + 1000) = 500500,
2
which can be checked directly only with considerable eﬀort. SN =
4.2.2 Geometric series
Equation (4.1) is a particular example of a geometric series, which has the
characteristic that the ratio of successive terms is a constant (one-half in this
case). The sum of a geometric series is in general written
SN = a + ar + ar 2 + · · · + ar N−1 =
N−1
ar n ,
n=0
where a is a constant and r is the ratio of successive terms, the common ratio. The
sum may be evaluated by considering SN and rSN :
SN = a + ar + ar 2 + ar 3 + · · · + ar N−1 ,
rSN = ar + ar 2 + ar 3 + ar 4 + · · · + ar N .
If we now subtract the second equation from the ﬁrst we obtain
(1 − r)SN = a − ar N ,
and hence
SN =
a(1 − r N )
.
1−r
117
(4.3)
SERIES AND LIMITS
For a series with an inﬁnite number of terms and |r| < 1, we have limN→∞ r N = 0,
and the sum tends to the limit
a
S=
.
(4.4)
1−r
In (4.1), r = 12 , a = 12 , and so S = 1. For |r| ≥ 1, however, the series either diverges
or oscillates.
Consider a ball that drops from a height of 27 m and on each bounce retains only a third
of its kinetic energy; thus after one bounce it will return to a height of 9 m, after two
bounces to 3 m, and so on. Find the total distance travelled between the ﬁrst bounce and
the Mth bounce.
The total distance travelled between the ﬁrst bounce and the Mth bounce is given by the
sum of M − 1 terms:
M−2
9
SM−1 = 2 (9 + 3 + 1 + · · · ) = 2
3m
m=0
for M > 1, where the factor 2 is included to allow for both the upward and the downward
journey. Inside the parentheses we clearly have a geometric series with ﬁrst term 9 and
common ratio 1/3 and hence the distance is given by (4.3), i.e.
M−1 9 1 − 13
M−1 ,
= 27 1 − 13
SM−1 = 2 ×
1
1− 3
where the number of terms N in (4.3) has been replaced by M − 1. 4.2.3 Arithmetico-geometric series
An arithmetico-geometric series, as its name suggests, is a combined arithmetic
and geometric series. It has the general form
SN = a + (a + d)r + (a + 2d)r 2 + · · · + [a + (N − 1)d] r N−1 =
N−1
(a + nd)r n ,
n=0
and can be summed, in a similar way to a pure geometric series, by multiplying
by r and subtracting the result from the original series to obtain
(1 − r)SN = a + rd + r 2 d + · · · + r N−1 d − [a + (N − 1)d] r N .
Using the expression for the sum of a geometric series (4.3) and rearranging, we
ﬁnd
rd(1 − r N−1 )
a − [a + (N − 1)d] r N
+
SN =
.
1−r
(1 − r)2
For an inﬁnite series with |r| < 1, limN→∞ r N = 0 as in the previous subsection,
and the sum tends to the limit
rd
a
+
.
(4.5)
S=
1 − r (1 − r)2
As for a geometric series, if |r| ≥ 1 then the series either diverges or oscillates.
118
4.2 SUMMATION OF SERIES
Sum the series
S =2+
5
11
8
+ 3 + ··· .
+
2 22
2
This is an inﬁnite arithmetico-geometric series with a = 2, d = 3 and r = 1/2. Therefore,
from (4.5), we obtain S = 10. 4.2.4 The difference method
The diﬀerence method is sometimes useful in summing series that are more
complicated than the examples discussed above. Let us consider the general series
N
un = u1 + u2 + · · · + uN .
n=1
If the terms of the series, un , can be expressed in the form
un = f(n) − f(n − 1)
for some function f(n) then its (partial) sum is given by
SN =
N
un = f(N) − f(0).
n=1
This can be shown as follows. The sum is given by
SN = u1 + u2 + · · · + uN
and since un = f(n) − f(n − 1), it may be rewritten
SN = [ f(1) − f(0)] + [ f(2) − f(1)] + · · · + [ f(N) − f(N − 1)].
By cancelling terms we see that
SN = f(N) − f(0).
Evaluate the sum
N
n=1
Using partial fractions we ﬁnd
1
.
n(n + 1)
un = −
1
1
−
n+1 n
.
Hence un = f(n) − f(n − 1) with f(n) = −1/(n + 1), and so the sum is given by
SN = f(N) − f(0) = −
1
N
+1=
.
N+1
N+1
119
SERIES AND LIMITS
The diﬀerence method may be easily extended to evaluate sums in which each
term can be expressed in the form
un = f(n) − f(n − m),
(4.6)
where m is an integer. By writing out the sum to N terms with each term expressed
in this form, and cancelling terms in pairs as before, we ﬁnd
SN =
m
f(N − k + 1) −
k=1
m
f(1 − k).
k=1
Evaluate the sum
N
n=1
Using partial fractions we ﬁnd
un = −
1
.
n(n + 2)
1
1
.
−
2(n + 2) 2n
Hence un = f(n) − f(n − 2) with f(n) = −1/[2(n + 2)], and so the sum is given by
3 1
1
1
.
+
SN = f(N) + f(N − 1) − f(0) − f(−1) = −
4 2 N+2 N+1
In fact the diﬀerence method is quite ﬂexible and may be used to evaluate
sums even when each term cannot be expressed as in (4.6). The method still relies,
however, on being able to write un in terms of a single function such that most
terms in the sum cancel, leaving only a few terms at the beginning and the end.
This is best illustrated by an example.
Evaluate the sum
N
n=1
1
.
n(n + 1)(n + 2)
Using partial fractions we ﬁnd
un =
1
1
1
−
+ .
2(n + 2) n + 1 2n
Hence un = f(n) − 2f(n − 1) + f(n − 2) with f(n) = 1/[2(n + 2)]. If we write out the sum,
expressing each term un in this form, we ﬁnd that most terms cancel and the sum is given
by
1 1
1
1
.
SN = f(N) − f(N − 1) − f(0) + f(−1) = +
−
4 2 N+2 N+1
120
4.2 SUMMATION OF SERIES
4.2.5 Series involving natural numbers
Series consisting of the natural numbers 1, 2, 3, . . . , or the square or cube of these
numbers, occur frequently and deserve a special mention. Let us ﬁrst consider
the sum of the ﬁrst N natural numbers,
SN = 1 + 2 + 3 + · · · + N =
N
n.
n=1
This is clearly an arithmetic series with ﬁrst term a = 1 and common diﬀerence
d = 1. Therefore, from (4.2), SN = 12 N(N + 1).
Next, we consider the sum of the squares of the ﬁrst N natural numbers:
SN = 12 + 22 + 32 + . . . + N 2 =
N
n2 ,
n=1
which may be evaluated using the diﬀerence method. The nth term in the series
is un = n2 , which we need to express in the form f(n) − f(n − 1) for some function
f(n). Consider the function
f(n) = n(n + 1)(2n + 1)
⇒
f(n − 1) = (n − 1)n(2n − 1).
For this function f(n) − f(n − 1) = 6n2 , and so we can write
un = 16 [ f(n) − f(n − 1)].
Therefore, by the diﬀerence method,
SN = 16 [ f(N) − f(0)] = 16 N(N + 1)(2N + 1).
Finally, we calculate the sum of the cubes of the ﬁrst N natural numbers,
SN = 13 + 23 + 33 + · · · + N 3 =
N
n3 ,
n=1
again using the diﬀerence method. Consider the function
f(n) = [n(n + 1)]2
⇒
f(n − 1) = [(n − 1)n]2 ,
for which f(n) − f(n − 1) = 4n3 . Therefore we can write the general nth term of
the series as
un = 14 [ f(n) − f(n − 1)],
and using the diﬀerence method we ﬁnd
SN = 14 [ f(N) − f(0)] = 14 N 2 (N + 1)2 .
Note that this is the square of the sum of the natural numbers, i.e.
N 2
N
3
n =
n .
n=1
n=1
121
SERIES AND LIMITS
Sum the series
N
(n + 1)(n + 3).
n=1
The nth term in this series is
un = (n + 1)(n + 3) = n2 + 4n + 3,
and therefore we can write
N
(n + 1)(n + 3) =
N
n=1
(n2 + 4n + 3)
n=1
=
N
n=1
n2 + 4
N
n=1
n+
N
3
n=1
= 16 N(N + 1)(2N + 1) + 4 × 12 N(N + 1) + 3N
= 16 N(2N 2 + 15N + 31). 4.2.6 Transformation of series
A complicated series may sometimes be summed by transforming it into a
familiar series for which we already know the sum, perhaps a geometric series
or the Maclaurin expansion of a simple function (see subsection 4.6.3). Various
techniques are useful, and deciding which one to use in any given case is a matter
of experience. We now discuss a few of the more common methods.
The diﬀerentiation or integration of a series is often useful in transforming an
apparently intractable series into a more familiar one. If we wish to diﬀerentiate
or integrate a series that already depends on some variable then we may do so
in a straightforward manner.
Sum the series
S(x) =
x4
x5
x6
+
+
+ ··· .
3(0!) 4(1!) 5(2!)
Dividing both sides by x we obtain
x3
x4
x5
S(x)
=
+
+
+ ··· ,
x
3(0!) 4(1!) 5(2!)
which is easily diﬀerentiated to give
x2
x3
x4
x5
d S(x)
=
+
+
+
+ ··· .
dx
x
0!
1!
2!
3!
Recalling the Maclaurin expansion of exp x given in subsection 4.6.3, we recognise that
the RHS is equal to x2 exp x. Having done so, we can now integrate both sides to obtain
S(x)/x = x2 exp x dx.
122
4.2 SUMMATION OF SERIES
Integrating the RHS by parts we ﬁnd
S(x)/x = x2 exp x − 2x exp x + 2 exp x + c,
where the value of the constant of integration c can be ﬁxed by the requirement that
S(x)/x = 0 at x = 0. Thus we ﬁnd that c = −2 and that the sum is given by
S(x) = x3 exp x − 2x2 exp x + 2x exp x − 2x. Often, however, we require the sum of a series that does not depend on a
variable. In this case, in order that we may diﬀerentiate or integrate the series,
we deﬁne a function of some variable x such that the value of this function is
equal to the sum of the series for some particular value of x (usually at x = 1).
Sum the series
S =1+
2
4
3
+ 3 + ··· .
+
2 22
2
Let us begin by deﬁning the function
f(x) = 1 + 2x + 3x2 + 4x3 + · · · ,
so that the sum S = f(1/2). Integrating this function we obtain
f(x) dx = x + x2 + x3 + · · · ,
which we recognise as an inﬁnite geometric series with ﬁrst term a = x and common ratio
r = x. Therefore, from (4.4), we ﬁnd that the sum of this series is x/(1 − x). In other words
x
f(x) dx =
,
1−x
so that f(x) is given by
f(x) =
1
d x =
.
dx 1 − x
(1 − x)2
The sum of the original series is therefore S = f(1/2) = 4. Aside from diﬀerentiation and integration, an appropriate substitution can
sometimes transform a series into a more familiar form. In particular, series with
terms that contain trigonometric functions can often be summed by the use of
complex exponentials.
Sum the series
S(θ) = 1 + cos θ +
cos 2θ
cos 3θ
+
+ ··· .
2!
3!
Replacing the cosine terms with a complex exponential, we obtain
exp 2iθ
exp 3iθ
S(θ) = Re 1 + exp iθ +
+
+ ···
2!
3!
(exp iθ)2
(exp iθ)3
+
+ ··· .
= Re 1 + exp iθ +
2!
3!
123
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