# Let $\lim_{n\to\infty}(a_n)=\infty$ and $(b_n)_{n\in\mathbb N}$ be bounded. Prove that $\lim_{n\to\infty}(a_n+b_n)=\infty$.

## Question:

Let $$(a_n)_{n\in\mathbb N}$$ and $$(b_n)_{n\in\mathbb N}$$ be sequences with real values. Let $$\lim_{n\to\infty}(a_n)=\infty$$ and $$(b_n)_{n\in\mathbb N}$$ be bounded. Prove that $$\lim_{n\to\infty}(a_n+b_n)=\infty$$.

## Proof:

Since we know:

$$\lim_{n\to\infty}(a_n)=\infty\space$$ $$\Leftrightarrow\space$$ $$\forall K\in\mathbb R^+ \space \space\exists N_1\in\mathbb N$$: $$n\gt N_1 \space$$ $$\Rightarrow a_n\gt 2K.$$

and $$(b_n)_{n\in\mathbb N}$$ is bounded $$\Leftrightarrow\space$$ $$\exists K\in\mathbb R^+ \space$$ such that $$\space \forall n\in\mathbb N \space$$ $$|b_n|\le K.$$

Now taking $$N=N_1$$:

$$\Rightarrow$$ $$a_n\gt 2K$$ and $$|b_n|\le K$$

$$\Leftrightarrow$$ $$2K\lt a_n$$ and $$-K\le b_n \le K$$

$$\Rightarrow$$ $$a_n+b_n \gt 2K-K=K$$ $$\space \forall n\gt N$$

Hence $$\lim_{n\to\infty}(a_n+b_n)=\infty$$

This is an optional question from my Analysis I course, thought it would be a fun little proof. Be great if anyone could check what I have done -struggling to get to grips with these types of proofs.

• If $(b_n)$ is bounded, you want $\exists K\in \mathbb{R}^+$, such that $\forall n\in\mathbb{N}$, $|b_n|\leq K/2$. Certainly not for all $K$. Feb 18 '20 at 22:10
• Oh yeah oops copied it from paper -translation error. I'll fix that, thank you :) Feb 18 '20 at 22:20

Actually, asserting than $$(b_n)_{n\in\mathbb N}$$ is bounded means that there is some $$K>0$$ such that $$(\forall n\in\mathbb N):\lvert b_n\rvert\leqslant K$$. So, fix such a $$K$$. Take $$M>0$$; you want to prove that, for some $$N\in\mathbb N$$, you have$$n\geqslant N\implies a_n+b_n>M.$$But you know that there is some $$N\in\mathbb N$$ such that$$n\geqslant N\implies a_n>M+K$$and therefore, if $$n\geqslant N$$,$$a_n+b_n>M+K-K=M.$$

• Okay, that makes sense. So the problem was that I said "$\forall K\in\mathbb R^+$" instead of fixing K? Feb 18 '20 at 23:00
• That was the main problem. Also, it is false that $K-\frac K2=K$. Feb 18 '20 at 23:03
• Omg that's embarrassing :( Feb 18 '20 at 23:05
• I'll change that, thank you for pointing that out :) Feb 18 '20 at 23:08

Let $$M > 0$$ be a bound for $$b_n$$, i.e.

$$|b_n| , for $$n \in \mathbb{N}$$.

$$a_n-M

Let $$K>0$$.

Since $$a_n \rightarrow \infty$$:

For $$K+M >0$$ there is a $$n_0$$ s.t. for $$n \ge n_0$$

$$a_n > K+M$$;

Then $$a_n+b_n >a_n-M >K$$ , and we are done.

• Very nice! Better than mine :) Feb 18 '20 at 22:44