If {$a_n$} is unbounded, why then $\frac{a_n}{1+a_n}$ does not tend to 0 as $n \to \infty$? If {$a_n$} is unbounded, why then $\frac{a_n}{1+a_n}$ does not tend to 0 as $n \to \infty$?
I see this sentence on the website. And I don't understand that.
 A: If $a_n$ is unbounded, then $a_n/(1+a_n)$ tends doesn't tend to 0 as $n$ goes to infinity, because whenever $a_n$ is very very large (which it will sometimes be, if the sequence is unbounded), then $a_n/(1+a_n) \approx 1,$ since adding 1 to a very large number changes it very very little.
More precisely, assumed $a_n/(1+a_n)$ converges to 0. Then we could find $N$ so that if $n > N,$ then $|a_n/(1+a_n)| < 0.5,$ as that is what it means to converge.
But this inequality implies that $|a_n| < |0.5 + 0.5a_n|.$
Since $a_n$ is unbounded, infinitely many times we must have that $|a_n| > 2$ (as if it was only larger than 2 finitely many times, we could bound $a_n$). In particular, for some $n > N$ we must have $|a_n| > 2.$ But then either $a_n > 2$ or $a_n < -2.$
In the first case, our inequality $|a_n| < |0.5 + 0.5a_n|$ becomes $a_n < 0.5 + 0.5a_n$ which is absurd if $a_n > 2,$ since we can rewrite it as $a_n < 1.$
In the second case of $a_n < -2,$ our inequality becomes $-a_n < 0.5-0.5a_n.$ But we can rewrite this as $0.5a_n > -0.5,$ or $a_n > -1,$ contradicting $a_n < -2.$ So we can't converge to 0.
A: Because
$\dfrac{a_n}{1+a_n}
=1-\dfrac{1}{1+a_n}
$.
Since $a_n$ is unbounded,
$\dfrac{1}{1+a_n}$ has
$0$ as a limit point
(there may be more)
so
$\dfrac{a_n}{1+a_n}$
has 1 as a limit point.
There may be other limit points, but 1 is always a limit point.
A: Since $(a_n)$ is unbounded you can find a subsequence of $(a_n)$, say $(a_{k_n})$, such that $a_{k_n} \to \infty$ (do this inductively). Now if we let $b_n = \frac{a_n}{1+a_n} = 1 - \frac{1}{1+a_n}$ notice that $b_{k_n} \to 1$ since $\frac{1}{1+a_{k_n}} \to 0$. Therefore, $b_n \not\to 0$ since every subsequence of a convergent sequence converges to the same limit.
