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[數學,M1&M2] Convergence



Convergence

[隱藏]
Question

Prove that S_n  = Σ 1/n does not converge.

引用:
On9 Proof

1 + 1/2 + 1/3 + 1/4 + ... + 1/n
> 1 + 1/2 + 1/4 + .... = 2

So S_n has no upper bound.


Is there any better proof?

回覆 引用 TOP

A better proof: By Contrapositive

Assume S_n converges.

We have S_2n = 1 + 1/2 + 1/3 + .... + 1/n + 1/(n+1) + 1/(n+2) + ... + 1/(n+n)

Then S_2n - S_n
= 1/(n+1) + 1/(n+2) + ... + 1/(n+n)
> 1/(n+n) + 1/ (n+n) + ... 1/(n+n)
= 1/2

So lim [n->∞] (S_2n - Sn) >= 1/2 =/=0

This leads to contradiction that S_2n also converges to the same limit.

QED

[ 本帖最後由 Aeroblast 於 2019-5-17 03:33 AM 編輯 ]

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Question

Given:

lim [n->∞] a_n = a
and
lim [n->∞] b_n = b

where a < b

Prove that a_n < b_n for all positive integers n is WRONG.




回覆 引用 TOP

回覆 3#  Aeroblast 的帖子


Solution

Put a = 1/n and b = 1 - 1/n
f(n) = 1/n and g(n) = 1 - 1/n has an intersection point.
So a_n and b_n can be equal.

QED

[ 本帖最後由 Aeroblast 於 2019-5-17 05:28 AM 編輯 ]

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