2
$\begingroup$

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)?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – PrimeNumber Jan 3 '11 at 20:34
  • $\begingroup$ In (2), "infinitely many" would sound better than "infinite". $\endgroup$ – Hans Lundmark Jan 4 '11 at 7:26
5
$\begingroup$

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 $\sum a_n^2$.

$\endgroup$
  • $\begingroup$ 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 $a_{n}^2=1/n^2$? $\endgroup$ – daniel.jackson Jan 3 '11 at 20:40
  • $\begingroup$ 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. $\endgroup$ – timur Jan 3 '11 at 21:04
  • $\begingroup$ 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 $n$? $\endgroup$ – daniel.jackson Jan 4 '11 at 20:26
  • $\begingroup$ $f_n=g_n+O(h_n)$ 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 $n$. $\endgroup$ – timur Jan 4 '11 at 21:25
2
$\begingroup$

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 $.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.