# How do you use the limit comparison test to determine if Sigma sin(1/n) from [1,oo) is convergent or divergent?

Nov 20, 2016

Let ${a}_{n} = \sin \left(\frac{1}{n}\right)$ and ${b}_{n} = \frac{1}{n}$.

Then ${\lim}_{n \rightarrow \infty} {a}_{n} / {b}_{n} = {\lim}_{n \rightarrow \infty} \sin \frac{\frac{1}{n}}{\frac{1}{n}}$. There are multiple ways to approach this limit. The first is to replace the variables: as $n \rightarrow \infty , \frac{1}{n} \rightarrow 0$, so this can be rewritten as ${\lim}_{a \rightarrow 0} \sin \frac{a}{a}$, which is a fundamental limit: ${\lim}_{a \rightarrow 0} \sin \frac{a}{a} = 1$. So:

${\lim}_{n \rightarrow \infty} \sin \frac{\frac{1}{n}}{\frac{1}{n}} = {\lim}_{a \rightarrow 0} \sin \frac{a}{a} = 1$

Another way of going about the limit is to apply L'Hopital's rule, since it's in the indeterminate form $\frac{0}{0}$. So:

${\lim}_{n \rightarrow \infty} \sin \frac{\frac{1}{n}}{\frac{1}{n}} = {\lim}_{n \rightarrow \infty} \frac{- \frac{1}{n} ^ 2 \cos \left(\frac{1}{n}\right)}{- \frac{1}{n} ^ 2} = {\lim}_{n \rightarrow \infty} \cos \left(\frac{1}{n}\right) = \cos \left(0\right) = 1$

Either way, we see that ${\lim}_{n \rightarrow \infty} {a}_{n} / {b}_{n} = 1$, which is a positive, defined value.

According to the limit comparison test this tells us that $\sum {a}_{n}$ and $\sum {b}_{n}$ are either both convergent or both divergent.

Since ${b}_{n} = \frac{1}{n}$, we see that $\sum {b}_{n}$ is divergent (it's the harmonic series), so we can conclude that $\sum {a}_{n} = {\sum}_{n = 1}^{\infty} \sin \left(\frac{1}{n}\right)$ is also divergent.

Nov 21, 2016

${\sum}_{k = 1}^{\infty} \sin \left(\frac{1}{k}\right)$ diverges

#### Explanation:

sinx = x - x^3/(3!)+x^5/(5!)+cdots is an alternating series and
x-x^3/(3!) < sinx < x for $- \pi \le x \le \pi$ so

$\frac{1}{n} - \frac{1}{6 {n}^{3}} < \sin \left(\frac{1}{n}\right) < \frac{1}{n}$ but

${\sum}_{k = 1}^{\infty} \frac{1}{n} - \frac{1}{6 {n}^{3}}$ diverges because

${\sum}_{k = 1}^{\infty} - \frac{1}{6 {n}^{3}}$ converges ($\approx - 0.2$) and

${\sum}_{k = 1}^{\infty} \frac{1}{n}$ diverges