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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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    In (2), "infinitely many" would sound better than "infinite".2011-01-04

2 Answers 2

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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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    $f_n=g_n+O(h_n)$ 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 $n$.2011-01-04
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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).2011-01-03