Sequence concept(20pt)

(a) False. .e.g. s_n = (-1)^n
(b) True
since {s_n} converges implies lim (s_n)^2 = (lim s_n)(lim s_n) converges
(c) False e.g. s_n =(-1)^n , t_n=(-1)^(n+1) , then s_n + t_n =0 for all n.
(d) False e.g. s_n =(-1)^n , t_n=0 , then s_n + t_n =(-1)^n diverges.
(e) True
otherwise if both {s_n} and {t_n} converges, then lim( s_n + t_n) = (lim s_n) + (lim t_n) converges
(f) False e.g. s_n =0 for all n, then s_n t_n =0 for all n.


圖片參考：http://us.a2.yahoofs.com/users/45b36d09zd8f9dafd/1bf2scd/__sr_/e18dscd.jpg?phoV4sFB3wxEMUE.


2007-01-28 10:07:51 補充：
(e) 係 TRUE{sn + tn} diverges then at least one of {sn} or {tn} diverges. 否則，不是 " 至少 {sn} 或 {tn} 一組是發散的 " ，表示 " {sn} 和 {tn} 都是收斂的 " 。

2007-01-28 10:17:42 補充：
If both {sn} AND {tn} are convergent, then , say, lim(sn) =s and lim(tn) =t will imply lim(sn+tn) = lim(sn)+lim(tn) = s+t .This means that { sn+tn } converges to a number (s+t) and hence { sn+tn } is convergent.This will contradict to the assumption { sn+tn } diverges.

(a) False, e.g. take sn = (-1)^n
then the sequence {sn^2} = {1}, which converges to 1 but {(-1)^n} = {sn} diverges

(b) True, this comes from the contrapositive of the statement
' If {sn} converges, then {(sn)^2} converges '
(Note that 'p implies q' is logically equvalent to 'not q implies not p')

(c) No, take sn = log n, and tn = - logn
then both {sn} and {tn} diverges but {sn + tn} = {0}, which converge to 0.

(d) No, take sn = log n, and tn = 1
then {sn + tn} diverges but tn converges to 1
(in fact the statement become true if 'both' is replaced by 'at least one')

(e) Answered in (d)

(f) No, take sn = 1/n, and tn = ln n
then {sn} conberges to 0 and {tn} diverges but {sn * tn} = {(ln n)/ n}, which converge
to 0.