Verified answer

1. 0≤ |bn - b| ≤ an, lim(n-> ∞) an=0,

then lim |bn -b| =0 (by the squeeze thm.)

so, lim(n-> ∞) bn = b.

2. For all ε > 0

(1) {xn} is convergent

( let lim xn=L, assume L > = 0, or we can consider the sequence {-xn})

so for all ε> 0, there exists N> 0,

such that when n > N, | xn - L | < ε/2 holds, then L-ε/2 < xn < L+ε/2

(2) choose M > N > 0,

such that |x1+x2+...+xN|/M < ε/2 (since x1+...+xN is fixed)

when n > M , then

yn = { (x1+x2+...+xN)+[x(N+1)+...+x(n)] } / n

< (x1+...+xN) / n + [ (L+ε/2)+(L+ε/2)+...+(L+ε/2) ]/n

< |x1+...+xN| / M +[ (n-N)(L+ε/2) ] / n

< ε/2 + L + ε/2

so, yn < L+ ε

Similarly, yn > L -ε

thus |yn - L| < ε

hence, lim(n-> ∞) yn= L (by the definition of limit)

The inverse statement of problem 2 is not true. For example, xn=(-1)^n

then yn = (x1+x2+ ...+xn)/n = -1/n or 0/n -> 0, but {xn} does not exist.

第二題後面部份舉例的比較簡單,

取(xn) 為 1,-1,1,-1,.......

(xn)不收歛,但(yn)收歛到0.