Why does $a_n=\sqrt{n} + \sin(n)$ diverge?

Question

I know the sequence does not converge to a point, so it must diverge. It is bounded on the bottom by 0 and there is no upper bound. So does it diverge because it is not bounded or because it oscillates? Thanks.

Answer 1

The sequence $$ a_n = \sqrt{n} + \sin(n) $$ diverges because it grows without bound. For any given $M$ you can find an $n$ such that $a_n > M$. That's it.

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Now, it is bounded below, but not above. If you have a sequence that is bounded below and above, and if it is monotonic (i.e. strictly increasing or decreasing from a point) then it will be convergent. But your example is not bounded above.

Answer 2

For all $n \geq 1$ we have $\sqrt{n} + \sin n \geq \sqrt{n} - 1$; given any $M > 0$, we have $\sqrt{n} - 1 > M$ if $n > (M+1)^{2}$; but then $n > (M+1)^{2}$ only if $\sqrt{n} + \sin n > M$. This shows that the desired sequence diverges to infinity.

Answer 3

To expand on Thomas' answer we can see that $\sin(n) \in [-1,1]$ which is bounded both above and below and that $\sqrt{n}$ is unbounded. Then we can use the fact that sum of unbounded and bounded is unbounded.