Alternating Series and Leibniz`s Test Let a1,a2,a3,... be a sequence

Calculus Notes
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Alternating Series and Leibniz’s Test
Let a1 , a2 , a3 , . . . be a sequence of positive numbers. A series of the form
a1 − a2 + a3 − a4 + a5 − a6 + . . .
is said to be alternating because of the alternating sign pattern. (The series
−a1 + a2 − a3 + . . . is also alternating, but it is more reassuring to start
summation with a positive term.)
The partial sums Sn of an alternating series are evidently not monotone,
S1 > S2 ,
S2 < S3 ,
S3 > S4 ,
....
However, the subsequences of odd-numbered and of even-numbered partial sums
S1 , S3 , S5 , . . . ,
S2 , S4 , S6 , . . . ,
may exhibit monotonic behavior. In fact, S2n+1 and S2n are monotone if and only
if the original sequence a1 , a2 , a3 , . . . is monotone.
If convergent, an alternating series may not be absolutely convergent. For this
case one has a special test to detect convergence.
ALTERNATING SERIES TEST (Leibniz). If a1 , a2 , a3 , . . . is a sequence of
positive numbers monotonically decreasing to 0, then the series
a1 − a2 + a3 − a4 + a5 − a6 + . . .
converges.
It is not difficult to prove Leibniz’s test. Indeed, since
a1 ≥ a2 ≥ a3 ≥ . . . ,
we have
a1 ≥ a1 − a2 + a3 ≥ a1 − a2 + a3 − a4 + a5 ≥ . . .
a1 − a2 ≤ a1 − a2 + a3 − a4 ≤ a1 − a2 + a3 − a4 + a5 − a6 ≤ . . . ,
which means that S2n+1 is monotone decreasing and S2n is monotone increasing.
Also S2n+1 = S2n + a2n+1 > S2n for every n, implying that both sequences are
bounded and hence convergent. To see that S2n+1 and S2n converge to the same
limit, observe that limn→∞ (S2n+1 − S2n ) = limn→∞ a2n+1 = 0. Proof finished.
A couple of conclusions follow from the above argument. First,
S2n < S < S2n+1 ,
where S is the sum of the series. And second,
S − S2n < a2n+1 ,
S2n−1 − S < a2n .
2
EXAMPLE. The alternating harmonic series
∞
X
(−1)n+1
n=1
n
converges by Leibniz’s test. Indeed, the sign pattern is + − + − + . . . and, as
n → ∞, the term n1 monotonically decreases to 0.
To illustrate the error estimate, observe for instance that
1 1 1 1 1 1 1 1
1 − + − + − + − + ≈ .746
2 3 4 5 6 7 8 9
is larger than the true sum but by no more than 0.1.