8.1 Basic Concepts and Examples 455a =f 0, the geometric series converges iff lrl < 1, and for all such r,
oo a
.Z::::: arn = --. D
n=O 1 - rExamples 8.1.3 (a) The nonzero "constant" series, .Z::::: c = c + c + c + · · ·,
diverges if c =f 0, since its sequence of partial sums is {Sn} = {nc}, which
diverges.
00
(b) The nonzero "alternating constant" series .Z::::: (-l)nc = c-c+c-c+· · ·
n=O
diverges if c =f 0, since its sequence of partial sums is { c, 0, c, 0, c, 0, · · · }. D
You will recall from your first-year calculus course that much attention
was given there to various tests that can be used to determine whether a given
series converges or diverges. The next theorem is the most basic of these tests.
Theorem 8.1.4 (General Term Test) If .Z::::: an converges, then an --+ 0.
[Equivalently, if an f+ 0, then .Z::::: an diverges.]
n
Proof. Suppose .Z::::: an converges, say to S. Let Sn= .Z::::: ak. Then Sn--+ S.
k=l
Notice that Vn EN,
Sn+l = Sn + an+l, so
an+1 = Sn+1 - Sn.Thus, by the algebra of limits for sequences,
n..oo lim an+l = nlim ..oo Sn+l - n---too lim Sn
= S-S = 0.
Therefore, n-->oo lim an= 0. •
oo n
Example 8.1.5 A divergent series: .Z::::: ---
n=l 5n + 11
The general term test (Theorem 8.1.4) shows that this series diverges, since
n -+~=f.O D
5n + 11 5 ·The general term test gives a necessary condition for convergence,
not a sufficient condition. It cannot be used to prove that a series converges;
it can only be used to prove that a series diverges. Thus, it is best viewed as
a test for divergence, as in Example 8.1.5. The following example shows that a
series can diverge even though its general term converges to 0.