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$$\sum_1^\infty \sin \frac{1}{n}$$

So now I konw that to evaluate this I can just look at the limit as it reaches infinity. I see that it would result in 1 over 0, but it approaches 0 so I could that that it approaches sin0 which is 0 so doesn't this converge?

Paul the Pirate
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3 Answers3

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Hint: Show that for $0<x\le 1$ we have $x<2\sin x.$ Consequently, for integers $n\ge 1,$ we have $$\frac1n<2\sin\frac1n,$$ so $$\frac12\cdot\frac1n<\frac12\cdot2\sin\frac1n=\sin\frac1n.$$ Use Comparison test with (a multiple of) the harmonic series.

Cameron Buie
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  • I don't understand where sin went in the last part. – Paul the Pirate Jul 21 '13 at 16:15
  • I have rearranged things slightly to make it clearer. Does that help? – Cameron Buie Jul 21 '13 at 16:18
  • Not really, I don't understance where 1/2 and 1/n came from, why they are being multiplied or what is really going on. – Paul the Pirate Jul 21 '13 at 16:20
  • In order to use the comparison test with the series $$\sum_{n=1}^\infty\sin\frac1n,$$ we have to find another series $$\sum_{n=1}^\infty a_n$$ with positive $a_n$s such that either: (i) $a_n<\sin\frac1n$ for each $n$ and the series $\sum a_n$ diverges, or (ii) $\sin\frac1n<a_n$ for each $n$ and the series $\sum a_n$ converges. In the former case, we will have proved that $\sum\sin\frac1n$ diverges (since "adding up" smaller terms gets a divergent series, then of course we won't get a convergent series from larger terms). In the latter, we will have proved that $\sum\sin\frac1n$ converges ... – Cameron Buie Jul 21 '13 at 16:28
  • I understand the comparison test, I just don't really understand anything in your answer. Why those bounds on x, why n larger than 1? Why all that math after that? – Paul the Pirate Jul 21 '13 at 16:31
  • ...(since "adding up" larger terms gets a convergent series, then of course we won't get a divergent series from smaller terms). Those bounds on $x$ aren't tight. The inequality $x<2\sin x$ actually holds for some larger $x$-values, but that isn't relevant. All of the $n$ we're interested in are at least $1$, so that $$0<\frac1n\le 1$$ for all $n\ge 1$. Thus, the inequality $x<2\sin x$ for $0<x\le 1$ allows us (through the arithmetic shown) to prove that $\frac12\cdot\frac1n<\sin\frac1n$ for all $n\ge 1$, whence we can use the comparison test. – Cameron Buie Jul 21 '13 at 16:36
  • I guess I don't understand why it matters that $0 < \frac{1}{n} < 1$ and I still don't understand the other stuff. – Paul the Pirate Jul 21 '13 at 16:43
  • If we didn't know that $$0<\frac1n\le1$$ for all $n\ge 1,$ then we couldn't use the fact that $x<2\sin x$ for $0<x\le1$ to open the way for the comparison test. – Cameron Buie Jul 21 '13 at 16:51
  • Where does 2sin come in from? – Paul the Pirate Jul 21 '13 at 16:57
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    That just came from knowing that the graphs of $y=x$ and $y=\sin x$ closely coincide for $x$ near $0$, but that $y=\sin x$ is always below $y=x$ for positive $x$. I tried a few scale factors and $2$ did the trick. – Cameron Buie Jul 21 '13 at 17:09
  • Incidentally, for any scalar $\alpha>\frac1{\sin(1)},$ we have $x<\alpha\sin x$ for $0<x\le 1.$ $2$ just happens to be a nice, small scalar that works. – Cameron Buie Jul 21 '13 at 18:20
  • So you picked an $a_n$ that is smaller and diverged? So 1/n is smaller than sinx but you had to specify a range since it is a repeating valued function? – Paul the Pirate Jul 21 '13 at 20:52
  • Indeed, we found $0<a_n<\sin\frac1n$ such that $\sum a_n$ diverged. In particular, we found this to be true for $$a_n=\frac12\cdot\frac1n.$$ The fact that the sine function is periodic doesn't actually come into play here. I specified the range because the inequality $0<x<2\sin x$ isn't true everywhere, but is true where it matters. – Cameron Buie Jul 22 '13 at 00:28
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[I think you are referring to the divergence test in your post. I hope to clarify the source of confusion, but this is too long for a comment.]

Just to clarify the divergence test:

If $\lim_{n\to \infty} a_n \neq 0$, then $\sum_{n = 1}^\infty$ diverges.

Put differently (it it's contrapositive), if $\sum_{n = 0}^\infty a_n\;$ converges, then $\lim_{n\to \infty} a_n= 0$.

The converse is not necessarily true. For example, $a_n = \dfrac 1n$. It is certainly true that $\lim_{n\to \infty} \dfrac 1n = 0$. However, $$\sum_{n = 1}^\infty \frac 1n \;\;\text{diverges}$$

Indeed, in this example, the harmonic series $\sum_{n=1}^\infty \dfrac{1}{2n}$ will make for a good comparison, using the comparison test: knowing that $$\sum_{n = 1}^\infty \frac 1{2n}$$ diverges, and knowing that $$\dfrac 1{2n} \leq \sin\frac 1n$$, you can argue that your series must also diverge.

amWhy
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  • Is there any easy way to memorize all these contrapostive, counterpostive and whatever other variation of positives or falses there is? I think there are 3 or 4 tests I need to memorize that each have 4 sets of cases that are counter or contra or intra postive and false and such. – Paul the Pirate Jul 21 '13 at 16:20
  • It really does get confusing, Paul. It's good just to know a little logic: $P \implies Q$ is equivalent to $\lnot Q \implies \lnot P$, but is not equivalent to $Q\implies P$. – amWhy Jul 21 '13 at 16:22
  • The important thing to know about the divergence test: you are guaranteed that a series diverges if the limit as $n\to \infty$ of the general term $a_n$ does not equal zero. When the limit is equal to zero, it tells you nothing about whether the series $\sum a_n$ converges or diverges. – amWhy Jul 21 '13 at 16:24
  • Paul, is this any clearer for you now? Why not try the limit comparison test: $a_n = \sin\left(\dfrac 1n\right)$ and $b_n = \dfrac 1n$? Let me know if you've got this question tackled or not. – amWhy Jul 21 '13 at 22:51
  • This is just so much to learn in such a short amount of time I can't keep any of it straight without it all written down, which I won't have for the test. I understand that 1/n is larger than sin1/n so that 1/n diverges mean sin does as well. – Paul the Pirate Jul 21 '13 at 22:53
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    I agree, it can be a handful. You do seem to be learning an awful lot in a short time. Yes, you can be assured that $\sin(1/n)$ diverges. For this one, since we want to compare to a divergent $b_n$ that is less than $\sin{1/n},$ we can set $b_n = \frac 1{2n}$, which diverges, by the limit comparison test: $\lim\dfrac{\frac 1{2n}}{ \frac 1n} = 2$, hence $b_n = 1/(2n)$ diverges since $1/n$ diverges. So, since the series $\sum \frac 1{2n}$ diverges, and $\dfrac 12n \lt \sin(1/n)$, $\sin 1/n$ diverges. – amWhy Jul 21 '13 at 23:03
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Hint: $$x\in (0,1/2)\Rightarrow\sin x > x/2$$

TZakrevskiy
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