The Real And Complex Number Systems

So, by above sayings, we have prove that the convergence of the product for all x ∈ R.
8.43 (a) Let a n −1 n / n for n 1,2,... Show that1 a n diverges but that
∑ a n converges.
Proof: Clearly, ∑ a n converges since it is alternating series. Consider
2n
k2
2n
1 a k 1 −1k
k2
k
1 1 2
≤ 1 1 2
1 − 1 3
1 − 1 4
1 1 4
1 1 4
1 − 1
2n − 1
1 − 1
2n
1 1
2n
1 1
2n
1 1 2
1 − 1 4
1 −
1 2n
*
and note that
n
1 −
2k 1 : p n
k2
is decreasing. From the divergence of ∑ 1 , we know that p 2k n → 0. So,
That is,
k2
1 a k diverges to zero.
1 a k 0.
k2
(b) Let a 2n−1 −1/ n , a 2n 1/ n 1/n for n 1, 2, . . . Show that 1 a n
converges but ∑ a n diverges.
Proof: Clearly, ∑ a n diverges. Consider
and
2n
1 a k 1 a 2 1 a 3 1 a 4 1 a 2n
k2
31 a 3 1 a 4 1 a 2n
3 1 − 1
2 2
1 − 1
n n
2n1
1 a k 1 a 2 1 a 3 1 a 4 1 a 2n 1 a 2n1
k2
3 1 − 1 1 − 1
2 2
n n
By (*) and (**), we know that
1 a n converges
n
since k2
1 − 1
k k
converges.
1 − 1
n 1
8.44 Assume that a n ≥ 0 for each n 1, 2, . . . Assume further that
*
**