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.
4.2 SUMMATION OF SERIES
4.2.1 Arithmetic series
Anarithmetic 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
n=0
(a + nd).
Rewriting the series in the opposite order and adding this term by term to the original expression forSN, we find
SN=N
2 [a + a + (N− 1)d] =N
2(first term + last term). (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 witha = 1, d = 1 and N = 1000. Therefore, using (4.2) we find SN=1000
2 (1 + 1000) = 500500, which can be checked directly only with considerable effort.
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 + ar2+· · · + arN−1=
N−1
n=0
arn,
wherea is a constant and r is the ratio of successive terms, the common ratio. The sum may be evaluated by consideringSN andrSN:
SN=a + ar + ar2+ar3+· · · + arN−1, rSN=ar + ar2+ar3+ar4+· · · + arN. If we now subtract the second equation from the first we obtain
(1− r)SN=a− arN, and hence
SN= a(1− rN)
1− r . (4.3)
For a series with an infinite number of terms and|r| < 1, we have limN→∞rN= 0, and the sum tends to the limit
S = a
1− r. (4.4)
In (4.1),r =12,a =12, and soS = 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 theMth bounce.
The total distance travelled between the first bounce and theMth bounce is given by the sum ofM− 1 terms:
SM−1= 2 (9 + 3 + 1 +· · · ) = 2
M−2
m=0
9 3m
forM > 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.
SM−1= 2×9
1−1
3
M−1
1−13
= 27
1−1
3
M−1 , where the number of termsN 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)r2+· · · + [a + (N − 1)d] rN−1=
N−1
n=0
(a + nd)rn, and can be summed, in a similar way to a pure geometric series, by multiplying byr and subtracting the result from the original series to obtain
(1− r)SN=a + rd + r2d +· · · + rN−1d− [a + (N − 1)d] rN.
Using the expression for the sum of a geometric series (4.3) and rearranging, we find
SN= a− [a + (N − 1)d] rN
1− r +rd(1− rN−1) (1− r)2 .
For an infinite series with |r| < 1, limN→∞rN= 0 as in the previous subsection, and the sum tends to the limit
S = a
1− r+ rd
(1− r)2. (4.5)
As for a geometric series, if|r| ≥ 1 then the series either diverges or oscillates.
4.2 SUMMATION OF SERIES
Sum the series
S = 2 +5 2+ 8
22+11 23+· · · .
This is an infinite arithmetico-geometric series witha = 2, d = 3 and r = 1/2. Therefore, from (4.5), we obtainS = 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 n=1
un=u1+u2+· · · + uN. If the terms of the series,un, can be expressed in the form
un=f(n)− f(n − 1) for some functionf(n) then its (partial) sum is given by
SN=
N n=1
un=f(N)− f(0).
This can be shown as follows. The sum is given by SN=u1+u2+· · · + uN
and sinceun=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
1 n(n + 1).
Using partial fractions we find
un=−
1 n + 1−1
n
.
Henceun=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+ 1 = N N + 1.
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)
wherem 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 k=1
f(N− k + 1) −
m k=1
f(1− k).
Evaluate the sum
N n=1
1 n(n + 2).
Using partial fractions we find un=−
1
2(n + 2)− 1 2n
.
Henceun=f(n)− f(n − 2) with f(n) = −1/[2(n + 2)], and so the sum is given by SN=f(N) + f(N− 1) − f(0) − f(−1) =3
4−1 2
1
N + 2+ 1 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 writeunin 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
2(n + 2)− 1 n + 1+ 1
2n.
Henceun=f(n)− 2f(n − 1) + f(n − 2) with f(n) = 1/[2(n + 2)]. If we write out the sum, expressing each termunin this form, we find that most terms cancel and the sum is given by
SN=f(N)− f(N − 1) − f(0) + f(−1) =1 4+1
2
1
N + 2− 1 N + 1
.
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 firstN natural numbers,
SN= 1 + 2 + 3 +· · · + N =
N n=1
n.
This is clearly an arithmetic series with first terma = 1 and common difference d = 1. Therefore, from (4.2), SN=12N(N + 1).
Next, we consider the sum of the squares of the firstN natural numbers:
SN= 12+ 22+ 32+. . . + N2=
N n=1
n2,
which may be evaluated using the difference method. Thenth term in the series isun=n2, which we need to express in the formf(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 functionf(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)] =16N(N + 1)(2N + 1).
Finally, we calculate the sum of the cubes of the firstN natural numbers, SN= 13+ 23+ 33+· · · + N3=
N n=1
n3, again using the difference method. Consider the function
f(n) = [n(n + 1)]2 ⇒ f(n− 1) = [(n − 1)n]2,
for whichf(n)− f(n − 1) = 4n3. Therefore we can write the generalnth 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)] =14N2(N + 1)2. Note that this is the square of the sum of the natural numbers, i.e.
N n=1
n3=
N
n=1
n
2
.
Sum the series
N n=1
(n + 1)(n + 3).
Thenth term in this series is
un= (n + 1)(n + 3) = n2+ 4n + 3, and therefore we can write
N n=1
(n + 1)(n + 3) =
N n=1
(n2+ 4n + 3)
=
N n=1
n2+ 4
N n=1
n +
N n=1
3
=16N(N + 1)(2N + 1) + 4×12N(N + 1) + 3N
=16N(2N2+ 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 3(0!)+ x5
4(1!)+ x6 5(2!)+· · · .
Dividing both sides byx we obtain S (x)
x = x3 3(0!)+ x4
4(1!)+ x5 5(2!)+· · · , which is easily differentiated to give
d dx
S (x) x
=x2 0!+x3
1!+x4 2!+x5
3!+· · · .
Recalling the Maclaurin expansion of expx given in subsection 4.6.3, we recognise that the RHS is equal tox2expx. Having done so, we can now integrate both sides to obtain
S (x)/x =
x2expx dx.
4.2 SUMMATION OF SERIES
Integrating the RHS by parts we find
S (x)/x = x2expx− 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) = x3expx− 2x2expx + 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 ofx (usually at x = 1).
Sum the series
S = 1 +2 2+ 3
22+ 4 23+· · · . Let us begin by defining the function
f(x) = 1 + 2x + 3x2+ 4x3+· · · , so that the sumS = f(1/2). Integrating this function we obtain
f(x) dx = x + x2+x3+· · · ,
which we recognise as an infinite geometric series with first terma = 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
f(x) dx = x 1− x, so thatf(x) is given by
f(x) = d dx
x 1− x
= 1
(1− x)2. The sum of the original series is thereforeS = 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 2! +cos 3θ
3! +· · · . Replacing the cosine terms with a complex exponential, we obtain
S (θ) = Re
1 + expiθ +exp 2iθ
2! +exp 3iθ 3! +· · ·
= Re
1 + expiθ +(expiθ)2
2! +(expiθ)3 3! +· · ·
.
Again using the Maclaurin expansion of expx given in subsection 4.6.3, we notice that S (θ) = Re [exp(exp iθ)] = Re [exp(cos θ + i sin θ)]
= Re{[exp(cos θ)][exp(i sin θ)]} = [exp(cos θ)]Re [exp(i sin θ)]
= [exp(cosθ)][cos(sin θ)].