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! :)