Mathematical Methods for Physics and Engineering : A Comprehensive Guide

4.2 SUMMATION OF SERIES

Integrating the RHS by parts we find

S(x)/x=x^2 expx− 2 xexpx+2expx+c,

where the value of the constant of integrationccanbefixedbytherequirementthat
S(x)/x=0atx= 0. Thus we find thatc=−2 and that the sum is given by

S(x)=x^3 expx− 2 x^2 expx+2xexpx− 2 x.

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 variablexsuch that the value of this function is

equal to the sum of the series for some particular value ofx(usually atx=1).

Sum the series

S=1+

2

2

+

3

22

+

4

23

+···.

Let us begin by defining the function

f(x)=1+2x+3x^2 +4x^3 +···,

so that the sumS=f(1/2). Integrating this function we obtain
∫
f(x)dx=x+x^2 +x^3 +···,

which we recognise as an infinite geometric series with first terma=xand common ratio
r=x. Therefore, from (4.4), we find that the sum of this series isx/(1−x). In other words
∫
f(x)dx=

x

1 −x

,

so thatf(x)isgivenby

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+cosθ+

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!

+···

}

.