284 Solutions
Problems of Chapter 13
13.1
Let there be given two series
A = 2~] Uk and B — \ vk
with nonnegative terms.
(a) // Wfc < Vk, yk, the convergence of series B implies the convergence of
series A and the divergence of series A implies the divergence of series B.
Suppose that B is convergent. Let S = Yl™ vk be finite.
n n
^2 uk < ]P
v
k < S, n = 0,1,...
o o
thus partial sum of A is bounded, hence it is convergent.
Suppose that A is divergent. Thus its nth partial sum increases indefinitely
together with n.
n n
0 0
Thus, nth partial sum of B increases indefinitely together with n, too.
That is, B is divergent.
(b) // lrnifc-^oo ^ = a > 0, then series A and B are simultaneously convergent
and divergent.
limfc-»oo % = a > 0, vk > 0, V*. Then,
Ve > 0 3N 9 a - e < — < a + e, Vfc > N.
Vk
=>• Vk{a — e) < Uk < Vk(a + e). If B is convergent, so is ^^° Vk(a — e).
Thus, A is convergent by (a). If B is divergent, so is ^^° vk{a — e). Thus,
A is divergent by (a).
13.2
a) Eo°°^0 ~k- X
It is convergent for x = 0. Let us assume that x > 0.
I*±! = J*+i)i =* _• 0 as A-• oo.
uk * k + 1