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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 $0$.

enter image description here

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

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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 $n=50 \implies a_n=33$

enter image description here

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

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

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  • $\begingroup$ 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? $\endgroup$ – BMath Oct 27 at 14:00
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    $\begingroup$ @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 $\alpha>0$. $\endgroup$ – user Oct 27 at 14:03
  • $\begingroup$ @BMath And for$ f_n(x) = \frac{1}{n} x+\sin((\log n)^\alpha x)$ we need $\alpha>1$. $\endgroup$ – user Oct 27 at 14:09

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