Evaluate $\lim_{x\to 0}\frac{1}{1-\cos(x^2)}{\sum_{n=4}^\infty {n^5x^n} }$ I am trying to calculate the following limit:
$$\lim_{x\to 0}\dfrac{1}{1-\cos(x^2)}{\sum_{n=4}^\infty {n^5 x^n} }$$
In general, I don't really know how to deal with Limits of infinite summation, so I tried to turn the series into a function, but I couldn't find the function. 
Any help will be appreciated...
 A: Using L'Hôpital's Rule (twice) we get that 
$$
\lim_{x\to 0}\frac{x^4}{1-\cos (x^2)}=\lim_{y\to 0}\frac{y^2}{1-\cos y}=2.
$$
Then
$$
\sum_{n=4}^\infty n^5x^n=x^4\sum_{n=0}^\infty (n+4)^5 x^n=x^4 f(x),
$$
and the power series $f(x)=\sum_{n=0}^\infty (n+4)^5 x^n$ has radius of convergence $r=1$ (using ratio test) and $f(0)=4^5=1024$.
Thus
$$
\lim_{x\to 0}\frac{1}{1-\cos (x^2)}\sum_{n=4}^\infty n^5x^n=\lim_{x\to 0}\frac{x^4}{1-\cos x}\cdot  f(x)=2 \cdot f(0)=2048.
$$
A: The simplest way is to expand $\cos x^2$ into a Taylor series, then cancel out $x^4$ in the fraction.
See:
$$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$$
Therefore $$1-\cos(x^2) = \frac{x^4}{2!} - \frac{x^8}{4!} + \frac{x^{12}}{6!}-\cdots=\\=x^4\left(\frac{1}{2} - \frac{x^4 }{4!}+\frac{x^{12}}{6!} - \cdots\right)$$
As for the numerator, it equals $$4^5x^4 + 5^5 x^5 + 6^5 x^6\cdots=\\=x^4\left(4^5 + 5^5x + 6^5x^2+\dots\right).$$
Writing the fraction allows you to cancel out $x^4$, sending $x$ to $0$ tell you that all the factors stil containing $x$ will equal zero, leaving you with the solution.
A: As $\displaystyle\cos2y=1-2\sin^2y,1-\cos(x^2)=2\sin^2\dfrac{x^2}2$ which $O(x^4)$
As  for $|x|<1,$ and  setting $\displaystyle\frac1x=y$
$\displaystyle\lim_{n\to\infty} n^5x^n=\lim_{n\to\infty}\frac{n^5}{y^n}=\lim_{n\to\infty}\frac{5n^4}{y^n\ln y}$ (using L'Hospitals)
$\displaystyle\implies\lim_{n\to\infty} n^5x^n=\cdots=0$
So, each term $ n^5 x^n$ will vanish for $n\ge5$ 
