Calculus- Early Transcendentals, 2021a

346 Sequences and Series
Theorem 9.20: Series are Linear
Suppose that ∑a n and ∑b n are convergent series, and c is a constant. Then
1. ∑ca n is convergent and ∑ca n = c∑a n
2. ∑(a n + b n ) is convergent and ∑(a n + b n )=∑a n +∑b n .
Note that when c is non-zero, the converse of the first part of this theorem is also true. That is, if ∑ca n
is convergent, then ∑a n is also convergent; if ∑ca n converges then 1 c ∑ca n must converge.
On the other hand, the converse of the second part of the theorem is not true. For example, if a n = 1
and b n = −1, then ∑a n + ∑b n = ∑0 = 0 converges, but each of ∑a n and ∑b n diverges.
In general, the sequence of partial sums s n is harder to understand and analyze than the sequence of
terms a n , and it is difficult to determine whether series converge and if so to what. The following result
will let us deal with some simple cases easily.
Theorem 9.21: Divergence Test
If ∑a n converges then lim
n→∞
a n = 0.
Proof. Since ∑a n converges, lim s n = L and lim s n−1 = L, because this really says the same thing but
n→∞ n→∞
“renumbers” the terms. By Theorem 9.5,
lim
n→∞ (s n − s n−1 )= lim
n→∞
s n − lim
n→∞
s n−1 = L − L = 0.
But
s n − s n−1 =(a 0 + a 1 + a 2 + ···+ a n ) − (a 0 + a 1 + a 2 + ···+ a n−1 )=a n ,
so as desired lim a n = 0.
♣
n→∞
This theorem presents an easy divergence test: Given a series ∑a n , if the limit lim a n does not exist or
n→∞
has a value other than zero, the series diverges. Note well that the converse is not true: If lim a n = 0then
n→∞
the series does not necessarily converge.
Theorem 9.22: The n-th Term Test
If lim
n→∞
a n ≠ 0 or if the limit does not exist, then ∑a n diverges.
Proof. Consider the statement of the theorem in contrapositive form:
∞
If ∑ a n converges, then lim a n = 0.
n→∞
n=1
If s n are the partial sums of the series, then the assumption that the series converges gives us
lim
n→∞ s n = s