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


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.

  • 2
    $\begingroup$ 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 $K$. $\endgroup$
    – Dayton
    Feb 18, 2020 at 22:10
  • $\begingroup$ Oh yeah oops copied it from paper -translation error. I'll fix that, thank you :) $\endgroup$ Feb 18, 2020 at 22:20

2 Answers 2


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 $n\geqslant N$,$$a_n+b_n>M+K-K=M.$$

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    $\begingroup$ Okay, that makes sense. So the problem was that I said "$\forall K\in\mathbb R^+$" instead of fixing K? $\endgroup$ Feb 18, 2020 at 23:00
  • $\begingroup$ That was the main problem. Also, it is false that $K-\frac K2=K$. $\endgroup$ Feb 18, 2020 at 23:03
  • $\begingroup$ Omg that's embarrassing :( $\endgroup$ Feb 18, 2020 at 23:05
  • $\begingroup$ I'll change that, thank you for pointing that out :) $\endgroup$ Feb 18, 2020 at 23:08

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

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

$a_n-M <a_n+b_n;$

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.

  • $\begingroup$ Very nice! Better than mine :) $\endgroup$ Feb 18, 2020 at 22:44

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