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The problem:

Does the sequence {$\frac{n+sin(n)}{n}$} converge? Prove your answer.

Since this came from the ratio test section of my book, I figured that the way to prove this would be through the ratio test.

proof:

$\lim_{n\to \infty}$ |$\frac{(a_n + 1 )}{a_n} |= $$\frac {(n+1)+sin(n+1)}{(n+1)}$ * $\frac{n}{(n+sin(n))}$ = ??

My question is my approach okay for this question? If it a correct approach, can I get tips on how to simplify what I have so far? At the current moment, I don't even know how to simply this down so that I get one of the three results of the ratio test. This is meant to be done in an Intro to Real Analysis context.

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    ratio test is used to prove that series converge. To prove that the sequence converges show $\lim_\limits{n\to \infty} \frac {n+\sin n}{n} = 1$ – 2017-02-23
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    @DougM In my book there is a Lemma that is referred to as the "Ratio Test for Sequences". Where {$x_n$} is a sequence such that $x_n$ does not equal zero for all n and such that the limit L:= lim$_{n\to \infnty}$ $\frac{x_{n+1}}{x_n}$. Am I not allowed to use this? – 2017-02-23

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Note that $$-\frac{1}{n}\le \frac{\sin n}{n} \le \frac{1}{n}$$ As $-1 \le \sin n \le 1$. So by the squeeze theorem, we have that $$\lim_{n \to \infty} \frac{\sin n}{n}=0$$ As $\lim\limits_{n \to \infty} \frac{1}{n}=0$. Now note that $$\lim_{n \to \infty} \frac{n+\sin n}{n}=\lim_{n \to \infty}1+\lim_{n \to \infty}\frac{\sin n}{n}=1$$ So it converges to $1$.

You cannot use the ratio test as it is used to prove that a series converges.