Question 4.T.10: (Alternating Series Test) Let (xn) be a positive, decreasing...
(Alternating Series Test)
Let (x_{n}) be a positive, decreasing sequence whose limit is 0. Then the alternating series ∑^{\infty }_{n=1} (−1)^{n+1} x_{n} is convergent, and
\left|\sum\limits_{k=n+1}^{\infty}{(−1)^{k+1} x_{k}} \right| ≤ x_{n+1}. (4.2)
First we shall prove that the sequence of partial sums S_{n} converges, using ideas from Example 4.4.
S_{2n} = (x_{1} − x_{2}) + (x_{3} − x_{4} ) + · · · + (x_{2n−1} − x_{2n}).
Since x_{k} − x_{k+1} ≥ 0 for all k, the sequence S_{2n} is increasing. It is also bounded above because
S_{2n} = x_{1} − (x_{2} − x_{3}) − · · · − (x_{2n−2} − x_{2n−1}) − x_{2n} ≤ x_{1} (4.3)
for all n ∈ \mathbb{N}, hence \lim S_{2n} exists; call it S. On the other hand,
|S_{2n+1} − S| = |S_{2n} + x_{2n+1} − S|
≤ |S_{2n} − S| + |x_{2n+1}| .
In the limit as n → ∞, we obtain \lim S_{2n+1} = S. Therefore S_{n} → S.
The relation (4.2) is equivalent to
|S − S_{n}| ≤ x_{n+1},
that is, the error involved in approximating S by S_{n} does not exceed x_{n+1}. Now
S − S_{n} = (−1)^{n+2}(x_{n+1} − x_{n+2} + x_{n+3} − · · · )
|S − S_{n}| = x_{n+1} − x_{n+2} + x_{n+3} − · · · . (4.4)
The series (4.4) is of the same alternating type as S = x_{1}−x_{2}+x_{3}−· · · , the only difference being that it starts with x_{n+1} instead of x_{1}. By (4.3) we have
S = \lim S_{2n} ≤ x_{1},
and the corresponding upper bound on the series (4.4) is therefore x_{n+1}.
Thus |S − S_{n}| ≤ x_{n+1}.
