Convergence of an infinite sum

Question

Is it possible to use the comparison test for convergence in the following series?

$$\sum_{n=1}^\infty \sin \frac 1 n$$

The exercise says that I should find a linear function $f(x)$ that satisfies $f(x)<\sin(x)$ if $0<x<\pi/2$ but I don't understand how that would help.

Any ideas?

Thanks!

Answer 1

Note that $g(x) = x/2$ satisfies $\sin(x) > g(x)$ for all $x < \pi/{2}$. Moreover, we have that

$$ \sum_{n=1}^\infty g(1/n) = \sum_{n=1}^\infty \frac{1}{2n} $$

does not converge. What does this tell you?

Answer 2

$$ \sin\frac 1 n \geq \frac{1}{2n}, $$ so by comparison, the series your're asking about diverges.

The linear function involved is $x\mapsto x/2$. That function is ${}<\sin x$ if $x$ is positive and sufficiently small.

Answer 3

We shall use the following:

Fact. If $x\in [0,1]$, then $\sin x\ge ax$, where $a=\cos( 1)>0$.

Proof. For every $x\in [0,1]$ $$ \cos 1\le \cos x, $$ and integrating in $[0,x]$ we obtain $$ \int_0^x cos (1)\,dt\le \int_0^x\cos t\,dt, $$ and hence $$ ax= \big(\cos( 1)\big)\,x\le \sin x. \tag*{$\Box$} $$

Now, for every $n\in\mathbb N$, using the above Fact we ontin $$ \sin\frac{1}{n}\ge \frac{a}{n}, $$ and as $\sum_{n=1}^\infty \frac{a}{n}$ diverges, so does $\sum_{n=1}^\infty\sin\frac{1}{n}$, by virtue of the comparison test.

Answer 4

Another approach (otra idea): since $\;\sin\frac1n>0\;\;\forall\,n\in\Bbb N\;$ , we can use the limit comparison test:

$$\frac{\sin\frac1n}{\frac1n}\xrightarrow[n\to\infty]{}1\implies \sum_{n=1}^\infty\sin\frac1n\;\;\text{converges}\;\;\iff\sum_{n=1}^\infty\frac1n\;\;\text{converges}$$

and since the rightmost one is the harmonic series....