Evaluating $\lim\limits_{x \to 0} \frac1{1-\cos (x^2)}\sum\limits_{n=4}^{\infty} n^5x^n$ 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.
 A: 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 $\sum_{4\le n<\infty}n^5x^n ?$
A: 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 $2\cdot 4^5 = 2048$.
A: 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 $2 \cdot 4^5$.
A: 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.)
A: If you evaluate the series, you get
$$ -{\frac {{x}^{4} \left( 243\,{x}^{5}-1426\,{x}^{4}+3454\,{x}^{3}-4386
\,{x}^{2}+3019\,x-1024 \right) }{ \left( x-1 \right) ^{6} \left( 1-
\cos \left( {x}^{2} \right)  \right) }}
 $$
$$ \sim \frac{1024x^4}{1-(1-\frac{x^4}{2!}+\dots)}\sim \frac{1024 x^4}{x^4/2!}=2048. $$
