Mathematical Methods for Physics and Engineering : A Comprehensive Guide

SERIES AND LIMITS

Given that the series

∑∞

n=1^1 /ndiverges, determine whether the following series converges:

∑∞

n=1

4 n^2 −n− 3

n^3 +2n

. (4.13)

If we setun=(4n^2 −n−3)/(n^3 +2n)andvn=1/nthen the limit (4.12) becomes

ρ= lim

n→∞

[

(4n^2 −n−3)/(n^3 +2n)

1 /n

]

= lim

n→∞

[

4 n^3 −n^2 − 3 n

n^3 +2n

]

=4.

Sinceρis finite but non-zero and

∑

vndiverges, from (i) above

∑

unmust also diverge.

Integral test

The integral test is an extremely powerful means of investigating the convergence

of a series

∑

un. Suppose that there exists a functionf(x) which monotonically

decreases forxgreater than some fixed valuex 0 and for whichf(n)=un,i.e.the

value of the function at integer values ofxis equal to the corresponding term

in the series under investigation. Then it can be shown that, if the limit of the

integral

lim

N→∞

∫N

f(x)dx

exists, the series

∑

unis convergent. Otherwise the series diverges. Note that the

integral defined here has no lower limit; the test is sometimes stated with a lower

limit, equal to unity, for the integral, but this can lead to unnecessary difficulties.

Determine whether the following series converges:

∑∞

n=1

1

(n− 3 /2)^2

=4+4+

4

9

+

4

25

+···.

Let us consider the functionf(x)=(x− 3 /2)−^2. Clearlyf(n)=unandf(x) monotonically
decreases forx> 3 /2. Applying the integral test, we consider

lim

N→∞

∫N

1

(x− 3 /2)^2

dx= lim

N→∞

(

− 1

N− 3 / 2

)

=0.

Since the limit exists the series converges. Note, however, that if we had included a lower
limit, equal to unity, in the integral then we would have run into problems, since the
integrand diverges atx=3/2.

The integral test is also useful for examining the convergence of the Riemann

zeta series. This is a special series that occurs regularly and is of the form

∑∞

n=1

1

np

.

It converges forp>1 and diverges ifp≤1. These convergence criteria may be

derived as follows.