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Let $a_n$ be a sequence such that: $ a_{n+1}-a_n \ge \frac{1}{n}$. prove that $\lim_\limits{n \to \infty} a_n = \infty$.

SOLUTION: I want to prove this by disproving Cauchy's test. which means, I want to prove that:

There is an $\epsilon>0$ , such that for each $N$, there is $n>N$ and $p \in$ $\mathbb N$ such that:$|a_{n+p}-a_n| \ge \epsilon$.

but for some reason I couldn't find that $\epsilon$. the "$\frac{1}{n}$" made my job harder.

To conclude: I want to disprove that $a_n$ converges by disproving that its Cauchy's and from $ a_{n+1}-a_n \ge \frac{1}{n}$ we get that the sequence is monotonic increasing, thus, this will prove that: $\lim_\limits{n \to \infty} a_n = \infty$.

Am I on the right track? are all the conclusions that I've reached are right? How do I really complete this? any kind of help would be appreciated.

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  • $\begingroup$ Can you use the fact that $\sum_1^\inf\frac{1}{n}$ diverges ? $\endgroup$ – fjardon Nov 26 '15 at 14:10
  • $\begingroup$ No unfortuanately, Sums of series's are not in the material of the course im taking. $\endgroup$ – F1sargyan Nov 26 '15 at 14:12
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    $\begingroup$ Well, you're going to have to use the divergence of $\sum 1/n$ one way or another. If you're not allowed to use it, you'll just have to prove it yourself. Or you can find a proof here. $\endgroup$ – TonyK Nov 26 '15 at 14:31
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Proof the sequence is not Cauchy

Using what you know about: $a_{n+1} - a_n \geq \frac{1}{n}$ you have:

$|a_{n+p} - a_{n}| = |\sum_{i=0}^{p-1}(a_{n+i+1} - a_{n+i})| \\ = \sum_{i=0}^{p-1}(a_{n+i+1} - a_{n+i}) \quad \text{because} \quad a_{n+1} - a_n \geq \frac{1}{n} \geq 0\\ \geq \sum_{i=0}^{p-1}\frac{1}{n+i} \\ \geq \sum_{i=0}^{p-1}\frac{1}{n+p}$

You now have:

$|a_{n+p} - a_{n}| \geq \frac{p}{n+p}$

You also have: $lim_{p \to +\infty}\frac{p}{n+p} = 1$

This implies that whatever $n$ you choose, you will find a $p$ such that $\frac{p}{n+p} > \frac{1}{2}$

So if you choose $\epsilon = \frac{1}{2}$ you have your contradiction for the Cauchy.

Proof a monotonicaly increasing sequence can converge

As you stated that:

the sequence is monotonic increasing, thus, this will prove that: $lim_{n \to +\infty}{a_n}=+\infty$

This is not true, a counter example is: $1-\frac{1}{n}$

Indeed, there is a result that monotonically increasing bounded sequences do converge.

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  • $\begingroup$ No, but I wanted to say that after proving that my sequence diverges, the sequence you gave converges. $\endgroup$ – F1sargyan Nov 26 '15 at 14:52
  • $\begingroup$ I didn't really get this step: $$ \geq \sum_{i=0}^{p-1}|a_{n+i+1} - a_{n+i}| \\ \geq \sum_{i=0}^{p-1}\frac{1}{n+1} \\$$ $\endgroup$ – F1sargyan Nov 26 '15 at 15:01
  • $\begingroup$ @F1sargyan sorry, it's a typo, it is: $n+i$ not $n+1$ $\endgroup$ – fjardon Nov 26 '15 at 15:37
  • $\begingroup$ @F1sargyan my sequence converges but it doesn't have the property: $|a_{n+p}-a_{n}| \geq \frac{1}{n}$ $\endgroup$ – fjardon Nov 26 '15 at 15:40
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    $\begingroup$ @F1sargyan, well, you cannot choose $n$ to be infinity as $n \in \mathbb{N}$ and $+\infty \not\in \mathbb{N}$. But you can choose an arbitrary $n$, and still find $p > n$ such that $\frac{p}{n+p} > \frac{1}{2}$ for instance take: $p > 2\cdot n$ $\endgroup$ – fjardon Nov 26 '15 at 16:19
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Hint : you should consider $$\lim_{N\to\infty} \sum_{n=0}^N (a_{n+1}-a_n)=\lim_{N\to\infty} a_{N+1}-a_0$$ Edit : since sum of series isn't in the material : $$\begin{align} \lim_{N\to\infty} a_{N+1}-a_1 &=\lim_{N\to\infty} \sum_{n=1}^N (a_{n+1}-a_n) \\\\ &\ge \lim_{N\to\infty} \sum_{n=1}^N \frac{1}{n} \\\\ &\ge \lim_{N\to\infty} \sum_{n=1}^N \int_n^{n+1} \frac{1}{t}dt\\\\ &= \lim_{N\to\infty} \int_1^{N+1} \frac{1}{t}dt\\\\ &= \lim_{N\to\infty} \int_1^{N+1} \frac{1}{t}dt\\\\ &= \lim_{N\to\infty} \ln(N+1) = \infty \end{align}$$

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    $\begingroup$ Does this has to do with the fact that $ \sum_1^\infty\frac{1}{n} $ diverges? if yes then I can't solve it like that. $\endgroup$ – F1sargyan Nov 26 '15 at 14:20
  • $\begingroup$ @F1sargyan I've edited my answer $\endgroup$ – stity Nov 26 '15 at 14:33
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Hint: Look at a proof for the divergence of the harmonic series and apply to your situation by avoiding the word "series".

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  • $\begingroup$ Ok lets say that I know the proof for that, what I will get is : $a_{n+1}-a_1 \ge 1 + \frac{1}{2} +\frac{1}{3}++..++\frac{1}{n}. $ are you implying that now I can use this to disprove convergence by Cauchy's test or the definition of limit? $\endgroup$ – F1sargyan Nov 26 '15 at 14:38
  • $\begingroup$ A proof of divergence of the harmonic series shows that the right-hand side diverges. Hence, the left-hand side has to diverge, too. $\endgroup$ – gerw Nov 26 '15 at 14:40
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Hint Show that $$a_{2^{n+1}}-a_{2^n} \geq \frac{2^{n}}{2^{n+1}}$$

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$$a_{n+1}-a_1=\sum_{k=1}^{n}a_{k+1}-a_k \ge 1 + \frac{1}{2} +\frac{1}{3}++..++\frac{1}{n}\to \infty$$

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