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What is the following limit equal to and how do I prove it?

$$\lim_{x\to 0^+} \frac{1}{1-\cos(x^2)}\cdot \sum_{n=4}^\infty{n^5x^n} $$

I've tried l'hospital but it doesn't seem to help since I don't know what the series converges to at $x\to 0^+$

What am I missing?

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The power series for $\cos u$ is $$ \cos u=\sum_{n=0}^\infty (-1)^n\frac{u^{2n}}{(2n)!}. $$ Letting $u=x^2$ we have $$ \sum_{n=0}^\infty (-1)^n\frac{x^{4n}}{(2n)!}. $$ Thus $$ 1-\cos(x^2)=\sum_{n=1}^\infty (-1)^{n+1}\frac{x^{4n}}{(2n)!} $$ (the constant term cancels). So, your limit is $$ \lim_{x\to 0^+}\frac{\sum_{n=4}^\infty n^5x^n}{\sum_{n=1}^\infty (-1)^{n+1}\frac{x^{4n}}{(2n)!}} $$ The lowest power of the denominator and numerator are both $4$. You can now apply L'Hopital's four times and see what happens.

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    $\begingroup$ Or, alternatively, factor $x^4$ out of each and cancel. $\endgroup$ – Nick Peterson Oct 7 '13 at 12:54
  • $\begingroup$ Yes, do as @NicholasR.Peterson says. $\endgroup$ – Joe Johnson 126 Oct 7 '13 at 12:55
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Maybe you don't know what the series converges to; however, you do know how to take its derivatives! As long as you are inside the radius of convergence of a power series, the derivative of the sum is the sum of the derivatives.

Also, you can notice this: every term in the power series includes a positive power of $x$. (In fact, the lowest is $x^4$.) So, all terms tend to $0$ as $x\rightarrow0^+$.

Thus this is an indeterminate, "0/0"-type limit. The derivative of the power series is $$ \sum_{n=4}^{\infty}n^5\cdot n\cdot x^{n-1}=\sum_{n=4}^{\infty}n^6x^{n-1}=\sum_{n=3}^{\infty}(n+1)^6x^n, $$ while the derivative of the denominator is $$ \frac{d}{dx}\left[1-\cos(x^2)\right]=2x\sin(x^2). $$ Thus, by L'Hopital's rule, $$ \lim_{x\rightarrow0^{+}}\frac{1}{1-\cos(x^2)}\sum_{n=4}^{\infty}n^5x^n=\lim_{x\rightarrow0^{+}}\frac{1}{2x\sin(x^2)}\sum_{n=3}^{\infty}(n+1)^6x^n, $$ if this limit exists. Here, there is a common power of $x$ that you can cancel; then, try doing the same thing.

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  • $\begingroup$ @Clayton Whoops! Nice catch. $\endgroup$ – Nick Peterson Oct 7 '13 at 12:50
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First, notice this

$$ x \sim 0 \implies 1-\cos(x^2) \sim \frac{x^4}{2!}. $$

Now, we have

$$ \frac{1}{1-\cos(x^2)}\cdot \sum_{n=4}^\infty{n^5x^n} \sim \frac{2}{x^4}\cdot \sum_{n=4}^\infty{n^5x^n} = \frac{2}{x^4}\cdot \left( 4^5 x^4 + \sum_{n=5}^\infty{n^5 x^n } \right) $$ $$ \longrightarrow_{x\to 0} 2. 4^5+0 .$$

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  • $\begingroup$ @RonGordon: Thanks for the comment. $\endgroup$ – Mhenni Benghorbal Oct 7 '13 at 20:27
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Ideas:

$$f(x):=\frac1{1-x}=\sum_{n=0}^\infty x^n\implies f'(x)=\sum nx^{n-1}\;,\;f''(x)=\sum n(n-1)x^{n-2}\ldots$$

$$f^{(iv)}(x)=\sum(n-3)(n-2)(n-1)nx^{n-4}$$

But for example

$$f''(x)=\sum_{n=2}^\infty n^2x^{n-2}-\sum_{n=2}^\infty n x^{n-2}\implies$$

$$\sum_{n=2}^\infty n^2x^n=f''(x)+\frac1x\sum_{n=2}^\infty nx^{n-1}$$

and you can express the right side in an elementary way...etc.

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