TITLE: Evaluate $\lim_{n \to \infty} n^3 \left(\int_0^{\pi}\cosh(2\cos(x)) dx - \pi\sum_{k=0}^n \frac{1}{(k!)^2} \right)$
QUESTION [11 upvotes]: Would you kindly show me how to evaluate this limit?
$$\lim_{n \to \infty} n^3 \left(\int_0^{\pi}\cosh(2\cos(x)) dx - \pi\sum_{k=0}^n \frac{1}{(k!)^2} \right)$$

REPLY [2 votes]: $$\int_0^{\pi} \cosh(2\cos x)dx=\sum_{k=0}^{\infty}\dfrac {2^{2k}}{(2k)!}\int_0^\pi \cos^{2k}xdx=\sum_{k=0}^{\infty}\dfrac{\pi \cdot2^{2k}}{(2k)!}\prod_{j=1}^{\infty}\dfrac{2j-1}{2j}=\pi \sum_{k=0}^{\infty}\dfrac{1}{(k!)^2}$$ So, what you want is
 the limit of $\sum_{k=1}^{\infty}\dfrac{\pi n^3}{((n+k)!)^2} $
$$0\leq\sum_{k=1}^{\infty}\dfrac{\pi n^3}{((n+k)!)^2}\leq \dfrac{n^3}{(n!)^2}(1+\frac{1}{n^2} +\frac{1}{n^4}..)$$ Squeeze theorem and we get the answer.