COMPLEX INFINITE SERIES.
© 2004-2009. RAIMOND A. STRUBLE, PhD.
DRAFT COPY ONLY.
9/27/2009.
© Raimond A. Struble.
Send comments and correspondence to: Raimond A. Struble,
P. O. Box 50376, Raleigh, NC 27650-6376
and emails to: George.Moore4@va.gov
Problem: For a sequence of complex numbers, prove that
(a)
∑1∞ zn /
(z1 + z2 +...+ zn) converges,
if and only if, ∑1∞ zn
converges, provided the denominators are non-zero, and do not tend
to zero as n → ∞.
(b) ∑1∞ zn /
(zn + zn+1 + ...) diverges, provided that
the (convergent) denominators are non-zero.
SOLUTION BY PROPOSER.
For (a), one observes that (with
sn = z1 + z2 + ... + zn),
(1 - zn/sn) = sn-1/sn
for n > 2. Therefore, the following partial product
(with rk = zk/sk), telescopes into
Pn = (1 - r2)(1 - r3) ...
(1 - rn) = z1/sn
(and z1 ≠ 0).
So long as sn does not tend to zero as
n → ∞, then the infinite product,
P = limn→∞ Pn, requires
that the two series
∑2∞ rn
and ∑2∞ zn =
limn→∞ sn converge or diverge together.
For (b), one observes that (with
un = zn + zn+1 + ...),
(1 - zn/un) = un-1/un
for n > 2. Therefore, the following partial product
(with tk = zk/uk
and s = z1 + z2 + ...) telescopes into
Qn = (1 - t2)(1 - t3) ...
(1 - tn) = un/s,
and tends to zero as n → ∞. The infinite product,
Q = limn→∞ Qn then requires
that the series ∑2∞ tn =
∑2∞ zn / un
diverges. One might take note of the interesting case for
zn = 1 / 2n, where
zn / un = 1 / [1 + (n / (n+1))2
+ (n / (n+2))2 + ... )]
< 1 / [1 + n(n / (n+n))2
+ n(n / (2n+n))2 + ... ]
= 1 / [1 + n(1/22 + 1/32 + ...)]
< 1 / [1 + 0.36n],
so that (b) is NOT trivial.
Last updated: 9/27/2009, by Raimond A. Struble, PhD.