<s>[INST] 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...
 [/INST] 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.
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