The Real And Complex Number Systems

For case (i), since
1
n log nlog log n p ≥ 1 for n ≥ 3,
n log n
1
we know that the series diverges by the divergence of ∑ .
n3 n logn
For case (ii), we consider (choose n 0 large enough)
∑
jn 0
2 j
2 j log 2 j log log 2 j p 1
log 2 ∑ jn 0
≥ ∑
jn 0
1
jlog j p ,
1
jlog j log 2 p
then, by Cauchy Condensation Theorem, the series diverges since ∑ jn0
by using Cauchy Condensation Theorem again.
For case (iii), we consider (choose n 0 large enough)
∑
jn 0
2 j
2 j log 2 j log log 2 j p 1
log 2 ∑ jn 0
≤ 2 ∑
jn 0
≤ 4 ∑
jn 0
1
jlog j log 2 p
1
jlog j log 2 p
1
jlog j p ,
1
jlogj p
1
jlogj p
diverges
then, by Cauchy Condensation Theorem, the series converges since ∑ jn0
converges by using Cauchy Condensation Theorem again.
Remark: There is another proof by Integral Test. We write it as a reference.
1
Proof: It is easy to check that fx is continous, positive, and
xlogxlog logx p
decreasing to zero on a, where a 0 for each fixed p. Consider
dx
dy
a x log xlog log x p
log loga y p
which implies that the series converges if p 1 and diverges if p ≤ 1byIntegral Test.
1
(j) ∑ n3 log logn
log logn
log logn
1
Proof: Leta n
log logn for n ≥ 3andbn 1/n, then
log logn
a n
n 1
b n log log n
e −ylogy−ey
→.
So, by Limit Comparison Test, the series diverges.
(k) ∑ n1
1 n 2 − n
Proof: Note that
1 n 2 − n 1
1 n 2 n ≥ 1 for all n.
1 2 n
So, the series diverges.