Re: [微積] 一題高微
※ 引述《eric80520 (freejustice)》之銘言：
: 題目 Suppose that {ak} is a decreasing sequence of real numbers.
: ∞
: Prove that if Σak converges, then kak→0 as k→∞.
: k=1
: 我目前會的
: ∞
: since Σak converges,so ak→0 as k→∞
: k=1
: by {ak} is decreasing and →0 ,so {ak} is nonegative sequence
: 再來好像要分2k跟2k+1討論, 可是我不太清楚要怎麼做
: 可以教我嗎 謝謝
Since Σa converges, then a → 0 as k → ∞
k k
and since {a_k} is decreasing, then a ≧ 0 , for all k in N
k
hence ka = a + a + ... + a
2k 2k 2k 2k
≦ a + a + ... + a
k+1 k+2 2k
2k
= Σ a
j=k+1 j
∵ Σa converges
k
2k
=> |ka | ≦ | Σ a | → 0 as k → ∞
2k j=k+1 j
∴ ka → 0 as k → ∞
2k
on the other hand
0 ≦ (2k+1)a ≦ (2k+1)a = 2ka + a → 0 as k → ∞
2k+1 2k 2k 2k
=> (2k+1)a → 0 as k → ∞
2k+1
therefore, ka → 0 as k → ∞
k
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