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For which $a>0$ series $$\sum { \left(2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right) \right)^a } $$ $(n \in \mathbb N)$ is convergent?

My try:
From Taylor theorem I know that:$$a_{n}={ \left(2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right) \right)^a } = (\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}}))^{a}$$ Then I have: $$(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}}))^{a} \le (\frac{1}{n^{4}}+o(\frac{1}{n^{8}}))^{a}$$At this point, my problem is that if I had: $$(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}}))^{a} \le (\frac{1}{n^{4}})^{a}$$I could say that $0 \le a_{n} \le (\frac{1}{n^{4}})^{a}$ so for $a>\frac{1}{4}$ this series is convergent.
However in this task I have also $o(\frac{1}{n^{8}}))^{a}$ and I don't know what I can do with it to finish my sollution.

Can you help me?

MP3129
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4 Answers4

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It is much simpler to use equivalents, which can be found with Taylor's formula:

  • $2(1-\cos x)=x^2-\dfrac{x^4}{12}+o(x^4)$,
  • $\sin(\sin x)=\sin\Bigl(x-\dfrac{x^3}6+o(x^3)\Bigr)=\Bigl(x-\dfrac{x^3}6\Bigr)-\frac16\Bigl(x-\dfrac{x^3}6\Bigr)^{\!3}+o(x^3)=x-\dfrac{x^3}3+o(x^3)$, so $$x\sin(\sin x)=x^2-\frac{x^4}3+o(x^4).$$ Now, replacing $x$ with $\frac 1n$, we obtain $$2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right)=\frac1{n^2}-\frac1{12n^4}-\frac1{n^2}+\frac1{3n^4}+o\biggl(\frac1{n^4}\biggr)=\frac1{4n^4}+o\biggl(\frac1{n^4}\biggr)$$ so an asymptotic equivalent for the general term of the series is $$\left(2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left(\sin\frac{1}{n}\right) \right)^a\sim_\infty\frac1{4^a n^{4a}}.$$

Knowing that series (with positive terms) which have asymptotically equivalent general terms both converge or both diverge, can you conclude?

Bernard
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  • Why you have ignored $o(1/n^4)$ part? – MP3129 Mar 26 '19 at 22:12
  • Because I use an equivalent function. The equivalent is the first non-zero term in Taylor's formula. I made the calculations with the $o$ part, to check my calculations are correct, but when I have obtained the equivalent, I can drop it. – Bernard Mar 26 '19 at 22:16
  • Ok, but we only know that $\frac{o(1/n^4)}{1/n^4} \rightarrow 0$, we don't know nothing about $ o(1/n^4) \rightarrow ? $, have I right? – MP3129 Mar 26 '19 at 22:19
  • As $1/n^4\to 0$, $o(1/n^4)\to 0$ too, of course. The main part of the function is the equivalent. – Bernard Mar 26 '19 at 22:21
  • by $o(1/n^4)$ you mean notation o-small or o-big? – MP3129 Mar 26 '19 at 22:42
  • I use notation o-small and in my opinion if $o(x)/x \rightarrow 0 $ it doesn't mean that $o(x) \rightarrow 0 $. To clarify: I use peano rest in Taylor function – MP3129 Mar 26 '19 at 22:45
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    I'm sorry, but if $o(x)/x \to 0$, we have $|o(x)/x|<1$ for small $x$, so $0\le|o(x)|<|x|$. Anyway being $o(x)$ implies being $O(x)$. – Bernard Mar 26 '19 at 22:51
  • :O "Anyway being o(x) implies being O(x)" I didn't know that! A lot thanks to you! – MP3129 Mar 26 '19 at 23:20
  • can you tell me one more thing - why in $\sin\Bigl(x-\dfrac{x^3}6+o(x^3)\Bigr) $ you ignored $o(x^3)$ and write for example $ \Bigl(x-\dfrac{x^3}6\Bigr)$ instead of $ \Bigl(x-\dfrac{x^3}6 + o(x^3) \Bigr)$ – MP3129 Mar 27 '19 at 21:08
  • That is because it is grouped with the final $o(x^3)$. In practise, when we expand a composition of functions, we compose only the polynomial parts, truncating what has a degree greater than the order of expansion. – Bernard Mar 27 '19 at 21:46
2

The answer is indeed as you concluded $a > \frac{1}{4}$.

Note that (from the first line in your attempt)

$$\left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right) = \theta\left(\frac{1}{n^4} \right),$$

for $n$ sufficiently large, and that is all you need to conclude

$$\left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right)^{a} = \theta \left(\frac{1}{n^4} \right)^a = \frac{1}{n^{4a}},$$

as $\sum_{n=1}^{\infty} \theta \left(\frac{1}{n^4} \right)^a$ converges iff $a > \frac{1}{4}$.

To elaborate:

$$\left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right) = \theta\left(\frac{1}{n^4} \right),$$

which implies for some positive constants $C_1, C_2$

$$ \frac{C_1}{n^4} \le \left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right) \le \frac{C_2}{n^4} $$

which implies for positive $a$:

$$ \frac{C_1}{n^{4a}} \le \left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right)^a \le \frac{C_2}{n^4a} $$

However, $\sum_{n=1}^{\infty} \frac{C_1}{n^{4a}}$ diverges for all positive $a \leq \frac{1}{4}$ so if $a$ is positive then $a$ must satisfy $a > \frac{1}{4}$. On the other hand $\frac{C_2}{n^{4a}}$ converges for all $a > \frac{1}{4}$ so it suffices that $a > \frac{1}{4}$.

Can use a similar line of reasoning to show that $a$ cannot be nonnegative for the sum to converge.

Mike
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  • The OP already concluded this. They want to prove that this is the solution. – Peter Foreman Mar 26 '19 at 21:52
  • And a question that my answer also does not address -- How do we know that the series diverges for $a=\frac14$? – Mark Fischler Mar 26 '19 at 21:54
  • Just revised See my edit – Mike Mar 26 '19 at 21:54
  • @Mike why you estimate $\left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right)$ by $\theta\left(\frac{1}{n^4} \right)$? Only what we know about $o(\frac{1}{n^{8}})$ is that $ \frac{o(1/n^8)}{1/n^8} \rightarrow 0 $ when $1/n^8 \rightarrow 0$ (so $n \rightarrow \infty$ ) – MP3129 Mar 26 '19 at 22:05
  • @MP3129 because $\frac{o(1/n^8)}{1/n^4}$ goes to 0, as does $-\frac{\theta(1/n^6)}{1/n^4}$. The $\frac{1}{n^4}$-term dominates. – Mike Mar 26 '19 at 22:09
  • I suspect that you have totally right, kill me but I can't understand why we have certainty to claim that $1/n^4$ dominates.... – MP3129 Mar 26 '19 at 22:11
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The term that is $o(n^{-8})$ is smaller, for all sufficiently large $n$, than $Cn^{-8}$, for at least some well-chosen value of $C$.

Making use of that $C$, you can finish the proof by noting that for sufficiently large $n$, $\frac7{72n^6} > Cn^{-8}$ since for large enough $n$, $$n^2 > \frac{72C}{7}$$.

Mark Fischler
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Use that you can find constants $0<c, C$ such that $c/n^4 \leq 1/n^4 -7/(72 n^6)+ o(1/n^8) \leq C /n^4$. Then you use minorant and majorant test to conclude that the series over the middle one converges iff the series over $1/n^4$ converges.

Severin Schraven
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