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I need to determinate if the following series converges:

$$\sum_{n=1}^{\infty}\sin{\tan{\frac{1}{n}}}$$ I see that $\sum_{n=1}^{\infty}\tan{\frac{1}{n}}$ obviously diverges, so does $\sum_{n=1}^{\infty}\sin\frac{1}{n}$ but still can't figure how to use the comparison test for the series above.

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    $\begingroup$ $\sin x > x/2$ for small positive $x.$ $\endgroup$ – zhw. Jan 30 '17 at 9:25
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We have $$\sin y\geq \frac{2}{\pi}y$$ for $0\leq y\leq \pi/2$ (by concavity) and $$\tan x\geq x$$ for $0\leq x< \pi/2$. Combining these, we get $$\sin\tan x\geq \frac{2}{\pi} x$$ for $0\leq x\leq \arctan(\pi/2)\simeq 1.00388$. In particular,

$$\sin \tan\frac{1}{n}\geq\frac{2}{n\pi}\quad\text{for all }n\geq 1$$

and hence the series diverges by the comparison test

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Using $\sin \epsilon \sim \epsilon$ ant $\tan \epsilon \sim \epsilon$ as $\epsilon\to 0$ and the quotient comparison test: $$\lim_{n\to +\infty}\dfrac{\sin{\tan{\dfrac{1}{n}}}}{\dfrac{1}{n}}=\lim_{n\to +\infty}\dfrac{{\tan{\dfrac{1}{n}}}}{\dfrac{1}{n}}=\lim_{n\to +\infty}\frac{\dfrac{1}{n}}{\dfrac{1}{n}}=1\ne 0$$ $$\Rightarrow \sum_{n=1}^{\infty}\sin{\tan{\dfrac{1}{n}}}\sim \sum_{n=1}^{\infty}{{\dfrac{1}{n}}}\text{ (divergent).}$$

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As this is a series with positive terms, you can use equivalence: $ \sin u\sim_0 u$, so the series $\sin\tan\dfrac1n\sim_\infty\tan \dfrac1n$, and equivalentseries with positive terms both converge or both diverge.

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  • $\begingroup$ Why should we use sin(x) -> x. I mean as n tends to infinity its obvious x tends to 0 and hence we can use it. But it wouldn't be good for previous terms when n is small. I am having this difficulty in getting the fact that when the nth term converges to something , the series converges. Because I think the previous terms should also affect. I hope I have explained my doubt properly. It has been a confusion for weeks now. $\endgroup$ – Shashaank Jan 30 '17 at 9:49
  • $\begingroup$ I didn't say $\sin x$ tend to $x$ – which is meaninglesss. I said it was equivalent near $0$, which has a very precise meaning in Asymptotic analysis. This allows to use the criterion of convergence /divergence for equivalent series. $\endgroup$ – Bernard Jan 30 '17 at 10:20
  • $\begingroup$ ok , thanks , got it ! $\endgroup$ – Shashaank Jan 30 '17 at 10:22
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The unction $f(x)=\sin{\tan{\dfrac{1}{x}}}$ is continuous, positive and decreasing in $[1,\infty)$, with integral test we have $$\sum_{n=1}^{\infty}\sin{\tan{\frac{1}{n}}}=\int_1^\infty\sin{\tan{\frac{1}{x}}}dx=\int_1^0\sin{\tan{u}}\frac{-1}{u^2}dx=\int_0^1\frac{\sin{\tan{u}}}{u^2}dx\geq\int_0^1\frac{1}{2u}dx=\infty$$

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