Spivak, Ch. 23, "Infinite Series", Problem 1(xiv): How do we show that $\sin{(1/n)}>\frac{1}{2n}$?

Question

Consider the infinite series

$$\sum\limits_{n=1}^\infty \sin{(1/n)}=\sin{(1)}+\sin{(1/2)}+\sin{(1/3)}+...\tag{1}$$

Figuring out if this converges is item (xiv) of problem 1 in Chapter 23 of Spivak's Calculus.

Here is my attempted solution

Let $f(x)=\sin{(1/x)}$. This function is positive and decreasing on $[1,\infty)$.

The sequence being summed is

$$a_n=\sin{(1/n)}=f(n)$$

If we write $\sin$ as an infinite series and integrate we find

$$\int\limits_0^{\infty} \sin{(1/x)}=\infty$$

Hence, by the integral test for infinite series, the series in (1) does not converge.

Then I looked at the solution manual and all it says is

Divergent, since

$$\sin{(1/n)}>\frac{1}{2n}\tag{2}$$

How do we know that (2) is true?

Answer 1

Using my favored $1,400^+$ years old approximation $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad \text{for} \qquad 0\leq x\leq\pi$$ So, making $x=\frac 1 n$ compute $$I_p=\int_1^p \frac{16 (\pi n-1)}{\pi n (5 \pi n-4)+4} \,dn$$ $$I_p=\frac{4 \left(2 \log (\pi p (5 \pi p-4)+4)+3 \tan ^{-1}\left(\frac{1}{4} (2-5 \pi p)\right)-2 \log (4+\pi (5 \pi -4))+3 \tan ^{-1}\left(\frac{1}{4} (5 \pi -2)\right)\right)}{5 \pi }$$

Expanding as a series for large $p$ $$I_p=\frac{16}{5 \pi }\log(p)+\frac{2 \left(-3 \pi +4 \log \left(\frac{5 \pi ^2}{4+\pi (5 \pi -4)}\right)+6 \tan ^{-1}\left(\frac{1}{4} (5 \pi -2)\right)\right)}{5 \pi }+O\left(\frac{1}{p}\right)$$ The constant term is quite small $(\sim -0.12)$.