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Prove that the sequence $a_n= 1+\dfrac{1}{2}+\dfrac{1}{3} + \ldots+\dfrac{1}{n}$ is not a cauchy sequence

The proof that my book has goes like this :

$a_ {2n }-a_n= \dfrac{1}{n+1} + \dfrac{1}{n+2}+\ldots + \dfrac{1}{2n} > n\cdot\dfrac{1}{2n}=\dfrac{1}{2}$

Then for every $n_0$ there exists $\epsilon=\dfrac{1}{2}$ , $p=n>n_0$ , $q=2n>n_0$ so that $|a_p-a_q|>\epsilon$

So its not cauchy.

I dont understand this. i thought that $p$ and $ q$ could be anything and that if i cant find any $p$ and $q$ so that $|a_p-a_q|>\epsilon$, then the sequence is not cauchy. For example, for the above proof, u can pick $p=2$ and $q=3$ and see that $|a_p-a_q|<\epsilon$ . Am i wrong? If yes, please explain it to me

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Compare your definition with this. –  S.B. Jul 15 '13 at 23:45
    
ah... now i get it -_- he proves that for every N there are at least one p and q so that |blabla|>ε .... pfft.. thanks , u can delete the post, stupid question –  Plom Jul 15 '13 at 23:52
    
Intuitively speaking, $\left< a_n\right>$ is Cauchy sequence means difference $|a_m-a_n|$ can be arbitrary small, if $m$ and $n$ are arbitary large, and this proof shows existence of a lower bound of $|a_m-a_n|$, it means $|a_m-a_n|$ is not arbitrary small, even $m$ and $n$ are arbitrary large. –  tetori Jul 16 '13 at 1:14

2 Answers 2

The point is that you don't have to find any $p$ and $q$.

For a given $\epsilon$ it must be always possible to find a $n_0$ ( this is what you have to find) such that for every $p$ and $q$ with $p,q \geq n_0$ the following is satisfied : $|a_p-a_q|<\epsilon$

In your special case such $n_0$ does not exist if we set $\epsilon:=1/2$, since the condition $|a_p-a_q|<\epsilon$ is not satisfied for every $p,q \geq n_0$. Infact if such a $n_0$ would exist then the condition would not be satisfied for $p:=n_0$ and $q:=2n_0$ as your book proofs.

I hope i could help a bit.

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yes, u helped a bunch, thanks for the answer :D –  Plom Jul 16 '13 at 0:33

Here is a hint. Look at $$\sum_{k = 2^n}^{2^{n + 1} -1} {1\over k}.$$

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