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I need to prove or disprove these two statements:

If $a_n > 0 $ for all natural numbers $n$, and $a_n$ is not bounded from above, then $a_n → \infty$

If $a_n → +∞ ⇔$ for all $M > 0$, there exist infinitely many terms of $a_n$ larger than M.

For the first one, I was thinking of a contradiction. We assume that the sequence converges to a finite value, and use that to show that there is a contradiction with 'not bounded from above'. Using the fact that $a_n$ > 0, we know that every value must be strictly > 0.

For the second one, I understand the intuition and image in my head, but I can't formulate a sound proof.

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  • $\begingroup$ For the first one, try the sequence $a_n = 1$ for odd $n$, and $a_n = n$ for even $n$. $a_n$ is then unbounded and positive, but there are infinitely many elements in it that are 'away from $\infty$', so $a_n \not \to \infty$. Only one of the two implications in the second is true, since the backward implication is essentially the same as the first. The forward implication follows from the definition of $a_n \to \infty$. $\endgroup$ – stochasticboy321 Sep 16 '16 at 6:56
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The first is not true. As a counterexample to your intuition consider a sequence with odd-indexed terms all equal $1$ and terms with even indices being consecutive natural numbers: $$(a_n)=1, \color{red}1, 1, \color{red}2, 1, \color{red}3, 1, \color{red}4, 1, \color{red}5, 1, \color{red}6, 1, \color{red}7,\ldots$$ For a sequence to diverge to $+\infty$ it's not enough to be unbound and greater than zero – it must be greater than any arbitrarily chosen value (at least from some point).

The second follows directly from the aforementioned condition for divergence to $+\infty$:
for any $M$, almost all terms of $(a_n)$ (therefore: infinitely many terms) must be greater than $M$.

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1) $$a_n=\begin{cases}\frac{1}{n}&n\in2\mathbb N\\ n&n\in2\mathbb N+1\end{cases}$$

2) "$\Rightarrow $" is correct but not "$\Leftarrow$" (see example over).

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  • $\begingroup$ why is the backwards direction not correct? if the example you are citing is the one posted by CiaPan, then I don't see how that disproves it $\endgroup$ – Chumbawoo Sep 16 '16 at 19:45
  • $\begingroup$ As you can see, for all $M$, there is infinitely many terms s.t. $a_n>M$, but the sequence doesn't diverge to $+\infty $. $\endgroup$ – Surb Sep 17 '16 at 6:28

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