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The comparison test does not work on this, so I'm stuck trying to find a way to prove that it diverges. I know it definitely diverges. Any solutions?

Jon Doe
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7 Answers7

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Consider the points $p_n=(\cos n,\sin n),$ $ n=1,2,\dots $ on the unit circle. The $p_n$ march around the circle infinitely many times in steps of of arc length $1.$ Let $A$ denote the arc on the unit circle between the points $(1/\sqrt 2,1/\sqrt 2)$ and $(-1/\sqrt 2,1/\sqrt 2).$ The length of this arc is $\pi/2>1.$ It is impossible for our sequence $p_n$ to jump over $A,$ as the $p_n$ are marching around in steps of arc length $1.$ Thus $p_n\in A$ for infinitely many $n.$ For each such $n,$ we have $\sin n \ge 1/\sqrt 2.$ It follows that $(e^n\sin n)/n \ge e^n/(\sqrt 2\cdot n)$ for infinitely many $n.$ Thus the terms of our series do not approach $0,$ which shows the series diverges.

zhw.
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    (+1) and happy holidays! – Mark Viola Dec 19 '18 at 17:08
  • Thanks, this seems like a good answer but is beyond the scope of calculus 2... I could never think of this on my feet. – Jon Doe Dec 19 '18 at 17:08
  • @JonDoe Just curious … what part of this answer is beyond the scope of a "Calculus 2" course? There will be an integer contained in any interval that is larger than $1$. You know that $\pi/2>1$. So, in particular, there will be an integer in each of the intervals $[2\pi m+\pi/4,2\pi m +3\pi/4]$. And in such an interval the value of the sine of the contained integer will exceed $1/\sqrt2$. – Mark Viola Dec 19 '18 at 17:26
  • @MarkViola Randomly thinking of "oh wait there are intervals 2pim + pi/4, 2pim + 3pi/4 set bounds to sinx. Let me use them!" Such is the same with integrals where say you have to add and subtract tan(x) or something to integrate it. It's not part of a student's mindset, particularly when they are not a math major. – Jon Doe Dec 19 '18 at 17:29
  • @JonDoe And now that you've seen this argument, could you add this to your "tool kit" and apply similar reasoning to other problems? If so, then this answer was useful and you should consider up voting it and accepting it. – Mark Viola Dec 19 '18 at 18:02
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    @JonDoe I recast the proof slightly. Tell me if it helps. – zhw. Dec 19 '18 at 18:17
  • @zhw Yes, this explanation helped quite a bit. Basically the span of the jump of 1 radian is less than the interval of values for which sin(x) < 1/ root(2). Therefore every few rotations the value of sin(x) > 1/ root(2) meaning that the function at some point will be greater than 1 / root(2) so the comparison test holds true for that. Such is the same for other points like 1 / 2. Thank you. – Jon Doe Dec 19 '18 at 18:39
  • I guess the thing that confused me was the bounds of sin(x) since it's -1 < x < 1. I thought you couldn't discount the other part of this interval and had to do something like sandwich it in order to compare (which doesn't work since the sandwich test doesn't work for infinite values) – Jon Doe Dec 19 '18 at 18:43
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If $|\sin(n)\,|\le\sin\left(\frac12\right)$, then $|\cos(n)\,|\ge\cos\left(\frac12\right)$ and $$ \begin{align} |\sin(n+1)\,| &=|\cos(n)\sin(1)+\sin(n)\cos(1)\,|\\ &\ge\cos\left(\tfrac12\right)\sin(1)-\sin\left(\tfrac12\right)\cos(1)\\ &=\sin\left(\tfrac12\right) \end{align} $$ This means that either $|\sin(n)\,|\ge\sin\left(\frac12\right)$ or $|\sin(n+1)\,|\ge\sin\left(\frac12\right)$.

Since $e^n\gt1+n$, we know that $\frac{e^n}n\gt1$.

Therefore, either $\left|\frac{e^n\sin(n)}n\right|\ge\sin\left(\frac12\right)$ or $\left|\frac{e^{n+1}\sin(n+1)}{n+1}\right|\ge\sin\left(\frac12\right)$. That is, the terms do not tend to $0$.

robjohn
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Take the root test. We have $\limsup_{n\to\infty}(|e^n\sin(n)|)^{1/n} = e> 1$ so it diverges.

Ben Grossmann
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kmini
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  • This is a convenient approach, assuming that the asker is familiar with the notion of a $\limsup$. Also, the fact that $\sin(n)$ has $1$ as its limsup as $n \to \infty$ is non-trivial. – Ben Grossmann Dec 19 '18 at 16:24
  • @kmini How does one show that $\sqrt[n]{\sin(n)}\to 1$ as $n\to \infty$? – Mark Viola Dec 19 '18 at 16:25
  • I'm not familiar with limsup. Where does the / n go? – Jon Doe Dec 19 '18 at 16:47
  • @MarkViola: note that $\lim\sup\sqrt[n]{\sin(n)}=1$ does not mean the same as $\sqrt[n]{\sin(n)}\to 1$. – TonyK Dec 19 '18 at 17:10
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    @TonyK Yes, I know the difference, Tony. So, I'll write more precisely to appease you. How does one show that $\limsup_{n\to\infty}\sqrt[n]{\sin(n)}=1$? – Mark Viola Dec 19 '18 at 17:13
  • @MarkViola: If you know the difference, why did you post your first comment? Does $\sqrt[n]{\sin(n)}$ tend to $1$? If so, it's not obvious to me. – TonyK Dec 19 '18 at 19:07
  • @TonyK I was abbreviating and have elaborated to appease you with a follow up comment. But you have not answered the question "How does one show that $\limsup_{n\to\infty}\sqrt[n]{\sin(n)}=1$? . – Mark Viola Dec 19 '18 at 19:18
  • @MarkViola: I never claimed to have an answer to that question. Why won't you just concede that your original comment was careless? – TonyK Dec 19 '18 at 21:21
  • @TonyK Stand down. You're obfuscating the bigger issue. The point of the matter is that the posted answer is completely useless unless it shows that the $\limsup \sqrt[n]{\sin(n)}=1$. Inasmuch as it does not, the answer is useless. – Mark Viola Dec 19 '18 at 21:29
  • @MarkViola: What answer? I didn't post any answer here. – TonyK Dec 19 '18 at 21:30
  • I've modified the previous comment – Mark Viola Dec 19 '18 at 21:31
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$$ \lim_{n \rightarrow \infty} \frac{e^n\sin n}{n} \neq 0 $$ So the series diverges using the nth term test for divergence.

Ahmad Bazzi
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It suffices to show that $\sin(n)$ fails to approach $0$, which means that your series fails to converge by the $n$th term test (that is, the sequence of summands fails to approach $0$).

Some proofs of that $\sin(n)$ fails to converge to $0$ are given here.

I would suspect, however, that if you are learning about testing for convergence/divergence of series for the first time, you would not be expected to prove such a result. If it comes up, I would state without proof that $\sin(n)$ does not have a limit as $n\to \infty$, just as the function $f(x) = \sin(x)$ fails to have a limit as $x \to \infty$.

Ben Grossmann
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  • That is a very weak argument. For example, $\sin(\pi n)$ does have a limit as $n\to\infty$, even though $f(x)=\sin(\pi x)$ doesn't. – TonyK Dec 19 '18 at 16:55
  • @TonyK I agree. However, I think that proving that $\sin(n) \to 0$ is beyond the scope of the typical "calc II" course in which this problem has probably appeared. – Ben Grossmann Dec 19 '18 at 16:58
  • You mean proving $\sin(n)\not\to 0$. – TonyK Dec 19 '18 at 17:08
  • @TonyK right. I suppose, though, proving that $\sin(n) \to 0$ is beyond the scope of any course in mathematics – Ben Grossmann Dec 19 '18 at 17:09
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The series does not converge because the general term does not tend to zero. To prove this, it is enough to show that $\sin n$ does not converge. Here is an elementary way to do it: suppose that $\sin n\to x\in \mathbb R$.

Then, of course,

$\sin 2n\to x$ and $\sin (2(n+1))\to x.$

And so

$\sin n\to x\Rightarrow \cos 2n\to 1-2x^2\Rightarrow \cos (2(n+1))\to 1-2x^2$

But now,

$\sin 2=\sin(2(n + 1) - 2n)=\sin(2(n + 1))\cdot\cos 2n-\sin 2n\cdot \cos (2(n+1))\to 0 $, which is absurd.

Matematleta
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Assuming $\frac{e^n \sin n}{n}\to 0$ as $n\to +\infty$ we have $\left|\sin(n)\right|\leq \frac{n}{2e^n}\leq\sin\frac{1}{2^n}$ for any $n$ sufficiently large, hence $\left|\cos n\right|\geq \cos\frac{1}{2^n}$. On the other hand such assumptions lead to $$ \left|\sin(n+1)\right|\geq \left|\left|\sin(n)\right|\cos(1)-\left|\cos(n)\right|\sin(1)\right|\geq \frac{4}{5} $$ for any $n$ sufficiently large, contradiction.

Jack D'Aurizio
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