ia2sp10h3s

# We need two further constants let α = ∑ ∞ n =0 |

This preview shows pages 2–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: We need two further constants. Let α = ∑ ∞ n =0 | a n | . Let C ≥ 0 be such that | t n | ≤ C for all n . Such a C exists because the series ∑ ∞ n =0 b n converges. (Convergent sequences are bounded.) Since B is the limit of the sequence { t n } , we also have | B | ≤ C . Let ² > 0 be given. Because ∑ ∞ n =0 | a n | < ² , there is N such that ∞ X n = N +1 | a n | < ² 3 C . Then also fl fl fl fl fl n X k =1 a k- A fl fl fl fl fl = fl fl fl fl fl ∞ X k = n +1 a k fl fl fl fl fl ≤ ∞ X k = n +1 | a k | ≤ ∞ X k = N +1 | a k | < ² 3 C if n ≥ N . Because ∑ ∞ n =0 b n = B , there is M such that if n ≥ M , then | B- t n | < ²/ (3 α ). Now one sees that if n > N + M n X k =0 c k = a ( b + ··· + b n ) + a 1 ( b + ··· + b n- 1 ) + ··· + a n- 1 ( b + b 1 ) + a n b n = n X k =0 a k s n- k . If k ≤ n- M then n- k ≥ M . For k = 0 ,...,n- M we write s n- k = B + ( s n- k- B ) and | s n- k- B | < ² . Now n- M > N , so ∑ n k = n- M +1 | a k | < ² . 2 Thus fl fl fl fl fl n X k =0 c k- AB fl fl fl fl fl = fl fl fl fl fl n X k =0 a k s n- k- AB fl fl fl fl fl = fl fl fl fl fl n- M X k =0 a k s n- k- AB + n X k = n- M +1 a k s n- k fl fl fl fl fl ≤ fl fl fl fl fl n- M X k =0 a k s n- k- AB fl fl fl fl fl + n X k = n- M +1 | a k || s n- k | ≤ fl fl fl fl fl n- M X k =0 a k ( B + s n- k- B )- AB fl fl fl fl fl + n X k = n- M +1 | a k || s n- k | = fl fl fl fl fl ˆ n- M X k =0 a k- A !...
View Full Document

{[ snackBarMessage ]}

### Page2 / 3

We need two further constants Let α = ∑ ∞ n =0 | a n |...

This preview shows document pages 2 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online