# Convergence of $\sum \limits_{n=1}^{\infty} (1-\frac{\sin a_{n}}{a_{n}})$ when $\sum \limits_{n=1}^{\infty} a_{n}$ converges

Let $\sum \limits_{n=1}^{\infty} a_{n}$ be a convergent series when $\forall n, a_{n}\neq 0$.

Does $\sum \limits_{n=1}^{\infty} (1-\frac{\sin a_{n}}{a_{n}})$ converge if:

1. $\forall n, a_{n}\gt 0$.
2. $a_{n}\lt 0$ for infinitely many $n$'s.

For (1) I tried looking at the Maclaurian series for $\sin x$:

$$|(1-\frac{\sin a_{n}}{a_{n}})|=|(1-\frac{1}{a_{n}}(a_{n}+R_{2}(a_{n}))|=|\frac{1}{6}a_{n}^2 \sin \xi| \leq a_{n}^2$$

I'm not sure if what I did above is correct but if it is then by the comparison test we get that the series converges. Is there a simpler argument for this? Maybe something that uses the fact that $\sin a_{n} \leq a_{n}$?

Regarding (2) I think that it can converge or diverge. I tried to guess and it seems that $a_{n}=\frac{(-1)^n}{\sqrt n}$ yields a divergent series according to Mathematica but I'm not sure why. How can I find one that converges or diverges without guessing (maybe again by using a Maclaurian series)?

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If you put $a_n = 1/n^2$ for (1), then $\sum_{n=1}^{\infty} \left(1- \frac{\sin a_n}{a_n} \right)$ converges. –  PEV Jan 3 '11 at 20:34
In (2), "infinitely many" would sound better than "infinite". –  Hans Lundmark Jan 4 '11 at 7:26

## 2 Answers

Since $\sum a_n$ converges, $a_n\to0$ as $n\to\infty$. The main thing to notice from the Taylor (or Maclaurin as you mentioned) series is that $\sin x=x-\frac16x^3+O(x^5)$ for small $x$. So we have $1-\frac{\sin a_n}{a_n}=a_n^2/6+O(a_n^4)$ for large $n$. Now one can see that the convergence of the series $\sum(1-\frac{\sin a_n}{a_n})$ is the same as the convergence of the series $\sum a_n^2$.

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Ok so I was pretty close in (1). If $\sum a_{n}$ converges and $a_{n}\gt 0$ then we know that $\sum a_{n}^2$ converges and then we're done. So for (2) I can say from the same arguments that if $a_{n}=\frac{(-1)^n}{\sqrt n}$ then the above series diverges since $a_{n}^2=1/n$ and if $a_{n}=\frac{(-1)^n}{n}$ the series converges since $a_{n}^2=1/n^2$? –  daniel.jackson Jan 3 '11 at 20:40
Yes you can, because the error ($\sum a_n^4$) is convergent. In other words, from (2) you cannot conclude anything, since examples of series with either behavior exist. –  timur Jan 3 '11 at 21:04
I understand your argument intuitively but I think I lack the justification why. When one writes $1-\frac{\sin a_n}{a_n}=a_n^2/6+O(a_n^4)$, the $O(a_n^4)$ means "an expression with powers larger than 4"? And if we look at it as a series then it tends to 0 for large $n$? –  daniel.jackson Jan 4 '11 at 20:26
$f_n=g_n+O(h_n)$ means the quantity $|f_n-g_n|/h_n$ is bounded as $n$ goes to infinity. In other words, $|f_n-g_n|\leq C h_n$ for some constant $C$ not depending on $n$. –  timur Jan 4 '11 at 21:25

For (1), you actually want $$(0 \le )\, 1 - \frac{{\sin x}}{x} \le x$$ for all sufficiently small $x>0$ ($x$ plays the role of $a_n \downarrow 0$). Indeed, it suffices to show $x - \sin x \leq x^2$, which follows straight by comparing derivatives (twice).

EDIT: For divergence in (2), consider a sequence $a_n = (-1)^n b_n$ where $b_n \downarrow 0$ monotonically, and note that $\sin x / x$ is an even function. Then, consider $1 - \sin x / x \geq x^3$ for all sufficiently small $x>0$. Finally, assume that $\sum \limits_{n = 1}^\infty {b_n^3 } = \infty$.

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In my answer to (2), consider the approach I used for (1). –  Shai Covo Jan 3 '11 at 21:55