convergence

(a)

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Crazyseries1.jpg

(b)

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Crazyseries2.jpg

Lemma: Σ a(n) converges==>a(n) --> 0 as n --> ∞
  n
Proof:Let s(n) =Σ a(k) --> L, hence a(n) = s(n) - s(n-1) --> L - L = 0, as n --> ∞
k = 1
(這叫做 n-term test。 比較嚴謹的 Proof 就要用 ε-δ argument。)

(a)

3n+1
Now a(n) = √----------.
16n - 3
 3n+1  3+1/n
Hencelim a(n) = lim √---------- = lim √---------- = √(3/16) = (√3)/4 ≠ 0
16n - 3  16 - 3/n
Hence, Σ a(n) diverges. (necessary condition is violated)


Lemma:y(n) > x(n) ==> lim y(n) ≥ lim x(n)

(b)

3n+13n
Now b(n) = √----------- > √--------- = {(√3)/4} (1/√n)
  16n^2 - 316n^2

Σ b(n) > {(√3)/4} Σ (1/√n), where Σ (1/√n) diverges. Hence Σ b(n) diverges.