What series do I compare $\displaystyle \sum_{n=1}^\infty \sin \frac 1n$ to when using the Limit Comparison test.

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- Dec 20th 2012, 01:39 PMMSUMathStdntLimit Comparison Test of the series of sin(1/n)
What series do I compare $\displaystyle \sum_{n=1}^\infty \sin \frac 1n$ to when using the Limit Comparison test.

- Dec 20th 2012, 01:44 PMPlatoRe: Limit Comparison Test of the series of sin(1/n)
- Dec 20th 2012, 02:27 PMSworDRe: Limit Comparison Test of the series of sin(1/n)
But that won't be sufficient, because for large n, $\displaystyle \sin{\frac{1}{n}} < \frac{1}{n}$

$\displaystyle \sum_{n=1}^\infty \frac{ 1}{n}$ diverges, but the series in the original post consists of smaller terms, so you can't use that as a direct comparison. You can however, use the fact that

$\displaystyle \sin{\frac{1}{n}} > \frac{1}{2n}$

and the fact that the below diverges:

$\displaystyle \sum_{n=1}^\infty \frac{ 1}{2n}$

In fact the 2 in the above series can be ANY number greater than 1. - Dec 20th 2012, 02:46 PMPlatoRe: Limit Comparison Test of the series of sin(1/n)

@SworD,**you did not read the question carefully**.

The question is aboutnot basic comparison.__limit__comparison

Some authors like Gillman call it*ratio comparison*.

$\displaystyle \lim _{n \to \infty } \frac{{\sin \left( {\frac{1}{n}} \right)}}{{\frac{1}{n}}} = 1$ - Dec 20th 2012, 08:13 PMhollywoodRe: Limit Comparison Test of the series of sin(1/n)
Yes, the (ordinary) comparison test is if $\displaystyle 0 \le a_n \le b_n$ where $\displaystyle \sum_{n=1}^\infty{b_n}$ is a convergent series, then $\displaystyle \sum_{n=1}^\infty{a_n}$ converges. Or if $\displaystyle 0 \le b_n \le a_n$ where $\displaystyle \sum_{n=1}^\infty{b_n}$ is a divergent series, then $\displaystyle \sum_{n=1}^\infty{a_n}$ diverges.

For the limit comparison test, if $\displaystyle \lim_{n \rightarrow \infty}\frac{a_n}{b_n}$ is finite and nonzero, then $\displaystyle \sum_{n=1}^\infty{a_n}$ converges if and only if $\displaystyle \sum_{n=1}^\infty{b_n}$ converges.

So you need to compare $\displaystyle \sin(1/n)$ to 1/n, since $\displaystyle \lim_{n \rightarrow \infty}\frac{\sin(1/n)}{1/n}=1$ is finite and nonzero. Of course, 1/2n or 35/87n would also work - the limits would still be finite and nonzero.

- Hollywood