500
8.4. ALTERNATING SERIES
exceeds S. Let
S 1 = P 1 .
(ii) Explain how we know that, regardless of the value of S 1 , we can find a
partial sum N 1
N 1 = −
m 1
k=1
1
2k
= −
1
2
−
1
4
−
1
6
− · · · −
1
2m 1
so that
S 2 = S 1 + N 1 ≤ S.
(iii) Explain how we know that, regardless of the value of S 2 , we can find a
partial sum P 2
P 2 =
n 2
k=n 1 +1
1
2k + 1
=
1
2(n 1 + 1) + 1
+
1
2(n 1 + 2) + 1
+ · · · +
1
2n 2 + 1
of the remaining positive terms of the alternating harmonic series so that
S 3 = S 2 + P 2 ≥ S.
(iv) Explain how we know that, regardless of the value of S 3 , we can find a
partial sum
N 2 = −
m 2
k=m 1 +1
1
2k
= −
1
2(m 1 + 1)
−
1
2(m 1 + 2)
− · · · −
1
2m 2
of the remaining negative terms of the alternating harmonic series so that
S 4 = S 3 + N 2 ≤ S.
(v) Explain why we can continue this process indefinitely and find a sequence
{S n } whose terms are partial sums of a rearrangement of the terms in the
alternating harmonic series so that lim
n→∞
S n = S.