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We have a sequence:

$$a_n=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$$

We need to show that it is rising and above unbounded.

So I did the following:

$$a_{n+1}=a_n+\frac{1}{n} $$

$$a_{n+1}>a_n $$

$$a_n+\frac{1}{n}>a_n $$

$$\frac{1}{n}>0$$

$$n>0$$

Which holds.

First question: Is the upper proof enough to satisfy that the sequence is rising

As far as I understand, I need to also now show that supremum doesn't exist.

So I have done the following:

$$ \sup{a_n}=\lim{a_n} = M $$

$$M>0$$

$$M>a_n$$ for every $n$

$$a_n+\frac{1}{n}<M$$

$$1<(M-a_n)n $$

Because $M>a_n$ we can divide without sign direction change

So we get:

$$\frac{1}{M-a_n}<n$$

Which I think isn't the end of the proof. I do not know how to continue.

VLC
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    for the first part, you wrote the proof backwards – J. W. Tanner May 28 '20 at 16:42
  • So if I understand correctly I had proven what I have asked ? – VLC May 28 '20 at 16:43
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    Your proof that the sequence is increasing is correct (maybe I'd emphasize that the statements you are using are all equivalent). Also, the sequence $(a_{n})$ is unbounded if and only if $\lim a_{n}=\infty$, since the sequence is increasing. Can you prove this last equality? – Johnny El Curvas May 28 '20 at 16:43
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    No one even tries to find a duplicate of this :[|||]: – metamorphy May 28 '20 at 16:45
  • @metamorphy The thing is that I found the same question but it wasn't proved in the same way as I wanted to do here – VLC May 28 '20 at 16:46

1 Answers1

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Here is a proof by grouping terms and using elementary inequalities (Due to Nicole Oresme).

For non-negative integers $j$ we have, $$\sum_{k=1}^{2^j} \frac{1}{2^j+k}\geq\sum_{k=1}^{2^j}\frac{1}{2^{j+1}}=\frac{2^j}{2^{j+1}}=\frac{1}{2}$$

Now, for $n\geq1$, $$\sum_{r=1}^{2^n}\frac{1}{r}=1+\sum_{j=0}^{n-1}\left( \sum_{k=1}^{2^j}\frac{1}{2^j+k}\right)\geq\left(1+\frac{n}{2}\right)=\frac{n+2}{2}$$

Equality holds for $n=1$.

Hence $a_{2^n}\geq\frac{n+2}{2}$. So the sequence $\{a_n\}$ grows unboundedly.

ShBh
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  • I have a follow up question: Could I also prove this way: $$f(x)=x$$ $$f(x)=x+1/n$$

    And then we get :

    $$1/n = 0$$

    Which doesn't hold for all $n$ and dus the sequence does not converge

     – VLC May 28 '20 at 16:50
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    what do you mean by $f$? – ShBh May 28 '20 at 16:52
  • A function of recursive sequence notation. I see if there is an intersection with the line $$y = x$$ and if there is none, it doesn't converge. Is this correct? – VLC May 28 '20 at 16:54
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    @BiliDebili Your proposed argument does not work, since $$b_n = \sum_{k=0}^n \frac{1}{2^k} = 2 - \frac{1}{2^n},$$ meaning that $b_n$ is bounded above by $2$ for all $n$; yet we also have $b_{n+1} > b_n$ for all $n$. Just because a sequence is increasing does not mean it is unbounded. – heropup May 28 '20 at 17:03