SERIES AND LIMITS
which is merely the series obtained by settingx= 1 in the Maclaurin expansion of expx
(see subsection 4.6.3), i.e.
exp(1) =e=1+1
1!
+
1
2!
+
1
3!
+···.
Clearly this second series is convergent, since it consists of only positive terms and has a
finite sum. Thus, since each termunin the series (4.7) is less than the corresponding term
1 /n! in (4.8), we conclude from the comparison test that (4.7) is also convergent.
D’Alembert’s ratio testThe ratio test determines whether a series converges by comparing the relative
magnitude of successive terms. If we consider a series
∑
unand setρ= lim
n→∞(
un+1
un)
, (4.9)then ifρ<1 the series is convergent; ifρ>1 the series is divergent; ifρ=1
then the behaviour of the series is undetermined by this test.
To prove this we observe that if the limit (4.9) is less than unity, i.e.ρ<1thenwe can find a valuerin the rangeρ<r<1 and a valueNsuch that
un+1
un<r,for alln>N. Now the termsunof the series that followuNare
uN+1,uN+2,uN+3, ...,and each of these is less than the corresponding term of
ruN,r^2 uN,r^3 uN, .... (4.10)However, the terms of (4.10) are those of a geometric series with a common
ratiorthat is less than unity. This geometric series consequently converges and
therefore, by the comparison test discussed above, so must the original series
∑
un. An analogous argument may be used to prove the divergent case when
ρ>1.
Determine whether the following series converges:
∑∞n=01
n!=
1
0!
+
1
1!
+
1
2!
+
1
3!
+···=2+
1
2!
+
1
3!
+···.
As mentioned in the previous example, this series may be obtained by settingx=1inthe
Maclaurin expansion of expx, and hence we know already that it converges and has the
sum exp(1) =e. Nevertheless, we may use the ratio test to confirm that it converges.
Using (4.9), we have
ρ= lim
n→∞[
n!
(n+1)!]
= lim
n→∞(
1
n+1)
= 0 (4.11)
and sinceρ<1, the series converges, as expected.