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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?

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  • $\begingroup$ 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
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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 $

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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 $a_n+b_n\gt M$...

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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 $n$.

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  • $\begingroup$ @ 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

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