The Real And Complex Number Systems

diverges, then ∑
where
By (2) and (4),
By (1) and (3),
−1log
k
k
also diverges. Consider
b k 1 m ... 1 m r , *
(1). log m N
(2). logm − 1 N − 1 log em − 1 N
(3). logm r N
(4). logm r 1 N 1 log mr1
e N
m r 1
e
m − 1 r 1 ≥ m if m is large enough.
2m ≥ r.
So, as k large enough ( m is large enough),
b k ≥ m r 1 r ≥ 3m m 1 by (*).
3
It implies that ∑−1 k b k diverges since b k does NOT tends to zero as k goes infinity.So,
k
−1log
we have proved that the series ∑ diverges.
k
(3) There is a good exercise by summation by parts, we write it as a reference.
Assume that ∑
k1
a k b k converges and b n ↗ with lim n→ b n . Show that b n ∑
kn
a k
converges.
Proof: First, we show that the convergence of ∑
k1
a k by Dirichlet Test as follows.
Since b n ↗ , there exists a positive integer n 0 such that as n n 0 ,wehaveb n 0. So,
1
we have
b nn0
is decreasing to zero. So
n1
∑ a kn0
k1
converges by Dirichlet Test.
∑a kn0 b kn0 1
b kn0
k1
For the convergence of b n ∑ kn
a k ,letn n 0 , then
b n ∑ a k ∑ a k b
b n k
b k
kn kn
and define c k a k b k and d k bn
b k
. Note that d k is decreasing to zero. Define
k
C k ∑ j1
c j and thus we have
m m
b n ∑ a k ∑ a k b
b n k
b k
kn kn
So,
m
∑C k − C k−1 d k
kn
m−1
∑ C k d k − d k1 C m d m − C n−1 d n .
kn
.