proof ($a_n+b_n$ converges) $\land$ ($a_n$ is bounded) then $b_n$ is bounded Tried to proof that if ($a_n+b_n$ converges) $\land$ ($a_n$ is bounded) then $b_n$ is bounded.
though I'm not sure I did everything correctly. would appreciate your advice.
so this is what I did:
given $a_n$ is bounded, there exist M>0 such that $|a_n| \le M$.
let's define K that $\lim_{n \to \infty}a_n+b_n \to K$
Now, $a_n+b_n$ converges, meaning $\lim_{n \to \infty}a_n+b_n = \lim_{n \to \infty}a_n + \lim_{n \to \infty}b_n \le M + \lim_{n \to \infty}b_n = K$
thus, $\lim_{n \to \infty}b_n = K-M$  while $(K-M)\in R$, thus $b_n$ converges and this is why it's bounded - as asked to prove.
so did I do everything ok?
 A: Hint:
Let $\lim a_n + b_n = L$ and $M$ be a bound on the $a_n$.
Then certainly there exists an $N$ such that 
$$n > N \ \Rightarrow \ |a_n + b_n - L | < 1 \ \Rightarrow \ |a_n + b_n| - |L| < 1$$
and for all such $n$ we also have that $|a_n| < M$.
Now try applying the triangle inequality to find a bound on $|b_n|$:
$$|b_n| = |a_n + b_n - a_n| \leq \cdots$$
A: By assumption there exists $x=\lim_n(a_n+b_n)$. In particular for each given $\varepsilon>0$ then $|a_n+b_n-x|<\varepsilon$ whenever $n\ge n_\varepsilon$. Now, if $|a_n|\le M$ for all $n$, this implies that for all $n\ge n_\varepsilon$ it holds
$$
|b_n|=|a_n+b_n-a_n| \le |a_n+b_n|+|a_n| \le |x|+\varepsilon+M
$$
It follows that for all $n$ we have
$$
|b_n| \le \max(b_1,\ldots,b_{n_\varepsilon-1},|x|+\varepsilon+M).
$$
A: Try this:
Since $a_{n}$ is bounded, there is some $M > 0$ so that for all $n$, we have $|a_{n}| < M$.
Now, since $a_{n} + b_{n}$ converges, then by definition that means there is some element $c$ so that for every $\epsilon > 0$, we can find a point $N$ in the sequence so that all later points $n \geq N$ satisfy $|a_{n} + b_{n} - c| < \epsilon$, i.e., $-\epsilon < a_{n} + b_{n} - c < \epsilon$.
Then $-\epsilon - a_{n} + c < b_{n} < \epsilon - a_{n} + c$ for all $n$.
Then if $K_{n} := \max\{ |\epsilon - a_{n} + c|, |-\epsilon - a_{n} + c| \}$, the above line gives us that $|b_{n}| \leq K_{n}$.  
But we also have $$K_{n} \leq \epsilon + |a_{n}| + |c| \leq \epsilon + M + |c| $$
Thus, $|b_{n}| \leq K_{n} \leq \epsilon + M + |c|$ for all $n \geq N$.  Then, just take the maximum of $|b_{1}|, \dots, |b_{N - 1}|, \epsilon + M + |c|$ and you get a finite number that bounds $|b_{n}|$ for all $n$.
