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I'm trying to solve this limit but I'm not sure how to do it.

$$\lim_{x \to 0} \frac1{1-\cos(x^2)}\sum_{n=4}^{\infty} n^5x^n$$ I thought of finding the function that represents the sum but I had a hard time finding it. I'd appreciate the help.

Nick Peterson
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matanc1
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  • Here is a technique to sum the series. – Mhenni Benghorbal Jul 20 '13 at 13:56
  • @Cortizol This edit was inappropriate - you changed the formula entirely. It is better in this sort of case to ask the OP for clarification, and ask them to edit their question themselves. – Nick Peterson Jul 20 '13 at 14:03
  • @NicholasR.Peterson Look at picture: $\cos x^2$, not $\cos^2 x$. So I edited. – Cortizol Jul 20 '13 at 14:04
  • Is it $\cos(x^2)$ or $\cos^2(x)$? – Mhenni Benghorbal Jul 20 '13 at 14:05
  • @Cortizol Ah, my apologies. I hadn't seen the post while the picture was still a part of it. I reverted. – Nick Peterson Jul 20 '13 at 14:05
  • @Cortizol: wrong! You see a contradiction - you ask the OP for clarification. Do not change the meaning/intent/content of a question, even if it is clear to you what it should be. – Ron Gordon Jul 20 '13 at 14:06
  • @RonGordon On the picture it was $\cos x^2$ (without brackets), then Gigili changed picture into TeX but with $\cos^2 x$, so I edited into $\cos x^2$. If some moderators can see full edit history that would be great. – Cortizol Jul 20 '13 at 14:08
  • When Shookie comes back tomorrow, she can tell us what was really intended... – GEdgar Jul 20 '13 at 14:09
  • @Cortizol: Shookie is the OP and is the one asking the question. By changing the content, you rendered my answer completely wrong. Perhaps I should have noticed the contradiction, but I typically answer the question posed. As an editor, it is OK for you to clean up the typography or fix typos, but please do not change content to the point where it can render someone's answer incorrect. – Ron Gordon Jul 20 '13 at 14:11
  • @RonGordon You don't get it. But we will see what OP is going to say. Until then. – Cortizol Jul 20 '13 at 14:18

5 Answers5

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HINT:

As $x\to0, |x|<1\implies \sum_{0\le n<\infty}x^n=\frac1{1-x}$ (Proof)

Differentiating wrt $x,$ $$\sum_{0\le n<\infty} nx^{n-1}=\frac1{(1-x)^2}\implies \sum_{0\le n<\infty} nx^n=\frac x{(1-x)^2}$$

Again differentiating wrt $x$ $$\sum_{0\le n<\infty}n^2x^{n-1}=\frac1{(1-x)^2}+\frac{2x}{(1-x)^3}\implies \sum_{0\le n<\infty}n^2x^n=\frac x{(1-x)^2}+\frac{2x^2}{(1-x)^3}$$

Can you continue the process to find $\sum_{0\le n<\infty}n^5x^n ,$ hence $\sum_{4\le n<\infty}n^5x^n ?$

lab bhattacharjee
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Of course finding a formula for the sum of the series is not needed to solve this.
$$\begin{align} \cos(x^2) &= 1 - \frac{x^4}{2} + O(x^8)\qquad\text{as }x \to 0 \\ \frac{1}{1-\cos(x^2)} &= 2 x^{-4} + O(1) \\ \sum_{n=4}^\infty n^5 x^n &= 4^5 x^4 + O(x^5) \\ \frac{1}{1-\cos(x^2)}\sum_{n=4}^\infty n^5 x^n &= 2\cdot 4^5 + O(x) \end{align}$$ and the limit is $2\cdot 4^5 = 2048$.

GEdgar
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You do not need to sum the series to determine the limit. As $x \to 0$, the sum $\sim 4^5 x^4$, while the denominator $= \sin^2{x} \sim x^2$ as $x \to 0$. Thus, the ratio $\sim 4^5 x^2$ as $x \to 0$, so the limit is $0$.

EDIT

An editor inappropriately changed the question after I posted this. The denominator is $1-\cos{x^2} \sim (x^2)^2/2$, so the limit is then $2 \cdot 4^5$.

Ron Gordon
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Note that the first term is $$\frac 1 {1-(1-x^4/2+O(x^8))} = \frac 2 {x^4}(1+O(x^4))$$ and that the second term is $$4^5 x^4(1+O(x))$$ (The question keeps changing; this is for the $$\cos x^2$$ denominator.)

not all wrong
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If you evaluate the series, you get

$$ -{\frac {{x}^{4} \left( 243\,{x}^{5}-1426\,{x}^{4}+3454\,{x}^{3}-4386 \,{x}^{2}+3019\,x-1024 \right) }{ \left( x-1 \right) ^{6} \left( 1- \cos \left( {x}^{2} \right) \right) }} $$

$$ \sim \frac{1024x^4}{1-(1-\frac{x^4}{2!}+\dots)}\sim \frac{1024 x^4}{x^4/2!}=2048. $$

Mhenni Benghorbal
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