Series with Taylor

Question

I'm trying to understand the solution of the following exercise: "Discuss the behavior of the series $$\sum _{n=1}^{\infty} (n-\sin n)(\dfrac{1}{n}-\sin \dfrac{1}{n})$$ Now, the solution on the textbook says that "because of Taylor's formula, we have that $$\sin x =x-\dfrac {x^3}{6}\cos {\xi} \qquad \text{with } 0<\xi<x$$ so $$\sin x \ge x-\dfrac{x^3}{6} \qquad \forall x \in [0,1]$$ And we can conclude that $$0 \le (n-\sin n)(\dfrac{1}{n}-\sin \dfrac{1}{n}) \le \dfrac{n+1}{6n^3} \le \dfrac{2n}{6n^3}=\dfrac{1}{3n^2}$$ So the series converges". I didn't understand second and third passages. I mean, why is only for $x \in [0,1]$?

Answer 1

I assume you understand why \begin{align} \sin x = x-\frac{x^3}{6}\cos\xi(x), \ \ 0<\xi(x)<x \end{align} where $x\in [0, \infty)$. In particular, we see that \begin{align} \sin x = x-\frac{x^3}{6}\cos \xi \ge x- \frac{x^3}{6} \end{align} since $\cos \xi \leq 1$. Again, this inequality holds for all $x \in [0, \infty)$ not just $x \in [0, 1]$. However, the solution only cares about it when $x<1$ because the author wants to apply this for $x=\frac{1}{n}$.

Using the above inequality for $x=n^{-1}$, we have that \begin{align} \left(\frac{1}{n}-\sin\frac{1}{n} \right) \leq \frac{1}{6n^3} \end{align} for all $n$. However, to deal with $n-\sin n$, the author doesn't use this inequality because it will given an overestimation (way over). Hence the author just use \begin{align} n-\sin n \leq n+|\sin n| \leq n+1 \end{align} which holds for all $n$. Thus, combining everything, you have the desired estimate \begin{align} (n-\sin n)\left(\frac{1}{n}-\sin\frac{1}{n} \right) \leq \frac{n+1}{6n^3} \leq \frac{1}{3n^2}. \end{align}

Answer 2

We have that

$$\frac{1}{n}-\sin\left(\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{n}+\frac{1}{3!n^3}-\frac{1}{5!n^5}+\cdots$$

$$=\frac{1}{3!n^3}-\frac{1}{5!n^5}+\cdots=\frac{1}{n^3}\left(\frac{1}{3!}-\frac{1}{5n^2}+\cdots\right)$$

Then

$$(n-\sin(n))\left(\frac{1}{n}-\sin\left(\frac{1}{n}\right)\right)=(n-\sin(n))\frac{1}{n^3}\left(\frac{1}{3!}-\frac{1}{5!n^2}+\cdots\right)$$

$$=\frac{1}{n^2}\left(\frac{1}{3!}-\frac{1}{5!n^2}+\cdots\right)-\frac{\sin(n)}{n^3}\left(\frac{1}{3!}-\frac{1}{5!n^2}+\cdots\right)$$

Taking the sum, we see

$$\sum_{n=1}^\infty (n-\sin(n))\left(\frac{1}{n}-\sin\left(\frac{1}{n}\right)\right)\leq \left|\sum_{n=1}^\infty (n-\sin(n))\left(\frac{1}{n}-\sin\left(\frac{1}{n}\right)\right) \right|$$

$$\leq \sum_{n=1}^\infty \left|(n-\sin(n))\left(\frac{1}{n}-\sin\left(\frac{1}{n}\right)\right)\right|$$

$$=\sum_{n=1}^\infty \left|\frac{1}{n^2}\left(\frac{1}{3!}-\frac{1}{5!n^2}+\cdots\right)-\frac{\sin(n)}{n^3}\left(\frac{1}{3!}-\frac{1}{5!n^2}+\cdots\right)\right|$$

$$\leq \sum_{n=1}^\infty \left|\frac{1}{n^2}\left(\frac{1}{3!}+\frac{1}{5!}+\cdots\right)\right|+\left|\frac{1}{n^3}\left(\frac{1}{3!}+\frac{1}{5!}+\cdots\right)\right|$$

$$=\sum_{n=1}^\infty\sum_{m=1}^\infty \frac{1}{(2m+1)!}\left(\frac{1}{n^2}+\frac{1}{n^3})\right)<\sum_{n=1}^\infty\sum_{m=1}^\infty \frac{2}{n^2(2m+1)!}$$

$$=\frac{1}{3} \pi ^2 (\sinh (1)-1)=0.576389$$

Thus, the sum is bounded and converges absolutely.