# Let $a_n$ equal the amount of roots of $f_n(x) = \frac{1}{n} x+\sin(x)$. Does $\sum_{n=1}^{\infty} \frac1{a_n}$ converge?

Let $$a_n$$ equal the amount of roots of $$f_n(x) = \frac{1}{n} x+\sin(x)$$. Does $$\sum_{n=1}^{\infty} \frac1{a_n}$$ converge?

For increasing $$n$$ the graph of $$f_n(x)$$ approaches the $$x$$-axis, the image below shows the first four iterations of $$f_n(x)$$. The amount of roots increases for bigger $$n$$. In particular, for $$n\to\infty$$ we see $$a_n \to \infty$$, because $$\lim_{n\to\infty} f_n(x) = \sin(x)$$. So the sequences $$\frac1{a_n}$$ converges to $$0$$.

Does the series converges, and if it does, is it there a specific limit?

Since for $$x>0$$ we have no roots when $$x>n$$

$$x>n \implies \frac{1}{n} x+\sin(x)> 0$$

and since every $$2\pi$$ cycle we have 2 roots then the amounts of roots $$a_n$$ can be estimated as

$$a_n\approx 1+2\cdot 2\cdot \frac{n}{2\pi} \sim n$$

and therefore the series diverges by limit comparison test with $$\sum \frac1n$$.

Here is a plot for $$n=50 \implies a_n=33$$

A more interesting problem could be the generalization with $$\alpha>0$$

$$f_n(x) = \frac{1}{n} x+\sin(n^\alpha x)$$

• Interesting generalization, going to the same procedure you might think $a_n \approx n^{1+\alpha}$, and $\sum \frac1{n^p}$ converges for $p>1$, however we approximate $a_n$, so can we get a clear bound for $\alpha$? Or does $\alpha > 0$ satisfy? – BMath Oct 27 at 14:00
• @BMath I think that the same argument can be used to show that $a_n \sim n^{1+\alpha}$ and therefore the series converges for any $\alpha>0$. – user Oct 27 at 14:03
• @BMath And for$f_n(x) = \frac{1}{n} x+\sin((\log n)^\alpha x)$ we need $\alpha>1$. – user Oct 27 at 14:09