I have the following question here.
How many terms do we need to sum for $\displaystyle \sum_{n=1}^{\infty} \sin\left( \frac{(-1)^{n+1}}{n^2} \right)$ so the error in the sum is less than $0.001$?
My attempt:
I know the alternating series estimation theorem says the following:
If $S=\displaystyle \sum_{n=1}^{\infty} (-1)^nb_n$ is the sum of an alternating series that satisfies:
i) $0\leq b_{n+1} \leq b_n$
ii) $\displaystyle \lim_{n \to \infty} b_n =0$
then $|R_n|=|S-S_n|\leq b_{n+1}$
Cool. So the official solutions does the following:
$\sin\left( \frac{1}{(n+1)^2} \right)<0.001$
$\frac{1}{(n+1)^2} < \arcsin(0.001)$
$n>30.622…$
$n\geq 31$
Why is this allowed though?
$\displaystyle \sum_{n=1}^{\infty} \sin\left( \frac{(-1)^{n+1}}{n^2} \right)$ isn't in the form $S=\displaystyle \sum_{n=1}^{\infty} (-1)^nb_n$ since the Sin term is in the way. Can someone explain how this actually works? Thanks!
Best Answer
As Robert Israel said, $\sin(\frac{(-1)^{n+1}}{n^2}) = (-1)^{n+1} \sin(1/n^2)$.
So the theorem applies: let $b_n = \sin(\frac{1}{n^2})$, and for $n>0$, we know that
To apply the theorem, you use $b_{n+1} = \sin(1/(n+1)^2)$.