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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.

Thomas
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Learner
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    because $\sqrt{n}$ diverges and $sin(n)$ which is between $-1$ and $1$ is negligible as the first term becomes large. – cr001 Nov 06 '15 at 13:27

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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.

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.

Thomas
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  • Thank you - great explanation! – Learner Nov 06 '15 at 13:32
  • The sequence $a_n$ is bounded below though, so this argument isn't sufficient to show that it doesn't converge since it isn't monotonic. – user1892304 Nov 06 '15 at 13:39
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    @user1892304: If it is monotonic and if it is bounded then it converges. You can use this fact to prove that it doesn't converge. It is true, however, that a convergent sequence is bounded, so if it is not bounded then it is not convergent. – Thomas Nov 06 '15 at 13:42
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    @Thomas Yes you're right, sorry. I was confusing the limit of a sequence and that of a function. – user1892304 Nov 06 '15 at 23:21
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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.

Yes
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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.

mathreadler
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