2
$\begingroup$

I am trying to prove the following problem

If $\lim\limits_{n\to\infty}a_n=+\infty$ and $\{b_n\}$ is a bounded sequence, then $\lim\limits_{n\to\infty}(a_n+b_n)=+\infty$

I have these definitions as tools;

Definition 1. The sequence $\{a_n\}\to \infty$ if $\forall \;M\in R,\;\;\exists\;n_0=n_0(M)\in N$ $\ni$ $$n\geq n_{0}\;\implies\;a_n>M.$$

Definition 2. The sequence $\{b_n\}$ is said to be bounded if there exists $M>0,$ $\ni$ $$|b_n|\leq M \;\forall \;n\geq 1.$$

Can anyone help me out?

$\endgroup$
  • $\begingroup$ Definition $2$ is wrong. The sequence is bounded if there exists $M$ such that $|b_n|\le M$ for all $n$. You should also have a go yourself and write out at work so far. $\endgroup$ – Jason Apr 12 '18 at 23:02
  • $\begingroup$ @ Jason: Thanks for that observation! I'll rectify it! $\endgroup$ – Omojola Micheal Apr 12 '18 at 23:04
2
$\begingroup$

We know that {b_n} is bounded therefore there exists a $B>0$ such that $-B< b_n< B $ for all $n\in \mathbb N$

Let $M>0$ be an arbitrary real number.

Since $\{a_n\}$ diverges to $\infty$, there exists some positive integer $N$, such that $$ n\ge N \implies a_n >M+B$$

Now if $ n\ge N $ we have $$a_n >M+B \implies a_n+b_n >M+B -B=M $$

Thus $ \{a_n + b_n \}$ diverges to $\infty $

$\endgroup$
1
$\begingroup$

Let $B$ be the bound for $(b_n)$. For any $M\gt0$ choose $n_0\in\mathbb N$ such that $n\ge n_0 \implies a_n\gt M+B$. Then $a_n+b_n\gt M$...

$\endgroup$
1
$\begingroup$

HINT: Rewriting part of your definitions:

  1. $\forall N\exists n_0\ \forall n\geq n_0\ a_n>N$.
  2. $\forall n |b_n|\leq M$.

It means that $a_n+b_n\geq N-M$ for large $n$.

$\endgroup$
  • $\begingroup$ @ Przemysław Scherwentke: Please, are you writing in terms of $a_n$ and $b_n$? $\endgroup$ – Omojola Micheal Apr 12 '18 at 23:24
  • $\begingroup$ @Mike Corrected. $\endgroup$ – Przemysław Scherwentke Apr 12 '18 at 23:26
  • $\begingroup$ @ Przemysław Scherwentke: That's good! Thanks! I think you still have to write $a_n$ in terms of $N$, i.e., $a_n>N$ $\endgroup$ – Omojola Micheal Apr 12 '18 at 23:27
  • $\begingroup$ @Mike Also corrected. $\endgroup$ – Przemysław Scherwentke Apr 12 '18 at 23:28
  • $\begingroup$ @ Przemysław Scherwentke: That's it! Thanks a lot for the hint! $\endgroup$ – Omojola Micheal Apr 12 '18 at 23:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.