Using ratiotest to check convergence of $\sum \sin(\frac{1}{n!})$.

Question

I'm preparing a class on the convergence of series and how to check it using de ratiotest. One of the exercises asks to determine the convergence of the series $$\sum \sin(\frac{1}{n!})$$ using the ratiotest.

I need to compute the following limit: $$\lim_{n \to \infty} \frac{\sin(\frac{1}{(n+1)!})}{\sin(\frac{1}{n!})}$$ but I have no idea how to do this: I tried using that $\sin(x) \leq x$, the sandwich theorem,...

I also can't recall any usefull geometric identity to split of a factor $\sin(\frac{1}{n!})$.

Any help would be very appreciated.

Answer 1

$$\dfrac{\sin\left(\dfrac{1}{(n+1)!}\right)}{\sin\left(\dfrac{1}{n!}\right)}=\dfrac{\sin\left(\dfrac{1}{(n+1)!}\right)}{\dfrac{1}{(n+1)!}}\times\dfrac{\dfrac{1}{n!}}{\sin\left(\dfrac{1}{n!}\right)}\times\dfrac{1}{n+1}$$

Answer 2

Note that\begin{align}\frac{\sin\left(\frac1{(n+1)!}\right)}{\sin\left(\frac1{n!}\right)}&=\frac{\frac{\sin\left(\frac1{(n+1)!}\right)}{\frac1{(n+1)!}}}{\frac{\sin\left(\frac1{n!}\right)}{\frac1{(n+1)!}}}\\&=\frac1{n+1}\times\frac{\frac{\sin\left(\frac1{(n+1)!}\right)}{\frac1{(n+1)!}}}{\frac{\sin\left(\frac1{n!}\right)}{\frac1{n!}}}.\end{align}Since$$\lim_{n\to\infty}\frac{\frac{\sin\left(\frac1{(n+1)!}\right)}{\frac1{(n+1)!}}}{\frac{\sin\left(\frac1{n!}\right)}{\frac1{n!}}}=\frac11=1,$$you know that$$\lim_{n\to\infty}\frac1{n+1}\times\frac{\frac{\sin\left(\frac1{(n+1)!}\right)}{\frac1{(n+1)!}}}{\frac{\sin\left(\frac1{n!}\right)}{\frac1{n!}}}=0.$$

Answer 3

Note that in this particular case, since $\sin(\frac 1{n!})>0$ you have directly from $\sin(x)\le x$

$$0\le\sum\limits_{n=0}^{\infty}\sin(\frac 1{n!})\le\sum\limits_{n=0}^{\infty}\frac 1{n!}=e$$

So the series is absolutely convergent (since $\nearrow$ and bounded above). But ok, if it is an exercise on the ratio test, so be it.