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)
wheremis an integer. By writing out the sum toNterms 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
2 n
]
.
Henceun=f(n)−f(n−2) withf(n)=− 1 /[2(n+2)],andsothesumisgivenby
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
2 n
.
Henceun=f(n)− 2 f(n−1) +f(n−2) withf(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) =