Convergence of $b_n=|a_n| + 1 - \sqrt {a_n^2+1}$ and $b_n = \frac{|a_n|}{1+|a_{n+2}|}$ here's my daily problem:
1) $b_n=|a_n| + 1 - \sqrt {a_n^2+1}$. I have to prove that, if $b_n$ converges to 0, then $a_n$ converges to 0 too. Here's how I have done, could someone please check if this is correct? I'm always afraid to square both sides.
$
\begin{align*}
0&=|a_n| + 1 - \sqrt {a_n^2+1}\\
& -|a_n| = 1 - \sqrt {a_n^2+1}\\
& a_n^2 = 1 - 2 *\sqrt {a_n^2+1} + a_n^2 + 1\\
& 2 = 2 *\sqrt {a_n^2+1}\\
& 1 = \sqrt {a_n^2+1}\\
& 1 = {a_n^2+1}\\
& a_n^2 = 0 \Rightarrow a_n=0\\
\end{align*}$
2) $b_n = \frac{|a_n|}{1+|a_{n+2}|}$
I have to prove the following statement is false with an example: "If $b_n$ converges to 0, then $a_n$ too."
I'm pretty lost here, any directions are welcome! I thought that would only converge to 0, if $a_n=0$. Maybe if $a_n >>> a_{n+2}$?
Thanks in advance! :)
 A: 1) One has $b_n=\dfrac{2|a_n|}{|a_n|+1+\sqrt{a_n^2+1}}\geqslant\dfrac{|a_n|}{1+|a_n|}\geqslant0$. 
Assume that $b_n\to0$. Then the inequality above shows that $\dfrac{|a_n|}{1+|a_n|}\to0$. If $\dfrac{|a_n|}{1+|a_n|}\leqslant\varepsilon$ with $\varepsilon\leqslant\frac12$, then $|a_n|\leqslant2\varepsilon$. Hence $|a_n|\to0$, which is equivalent to $a_n\to0$.
2) Try $a_n=2^{n^2}$.
Edit As regards the example suggested above for 2), it seems that a cul-de-sac was reached in the comments (see below, I cannot do any better to explain why what you say is false than what I already did), so let me try another example: let $a_n=n!$. 
Then $1+|a_{n+2}|\gt |a_{n+2}|=(n+2)(n+1)|a_n|$ hence $0\lt b_n\lt \dfrac{|a_n|}{|a_{n+2}|}=\dfrac1{(n+1)(n+2)}\to0$. One sees that $b_n\to0$ although $a_n\to+\infty$.
A: You seem to have a fundamental misconception regarding the difference between the limit of a sequence and an element of a sequence.
When we say $b_n$ converges to $0$, it does not mean $b_n = 0$ for all $n$. For instance $b_n = \frac{1}{n}$ is convergent to $0$, but there is no natural number $n$ for which $b_n = 0$.
In i) What you tried is ok (though the misconception above shows in the way you have written it), but are making some assumptions which need to be proved.
If we were to rewrite what you wrote, it would be something like,
If $a_n$ was convergent to $L$, then we would have that
$ 0 = |L| + 1 - \sqrt{L^2 + 1}$
and then the algebra shows that $L = 0$.
Using $a_n$ instead of $L$ makes what you wrote nonsensical.
Also, can you tell what assumption is being made here and needs justification?
For ii) Try constructing a sequence such that $\frac{a_{n+2}}{a_n} \to \infty$.
