# Evaluate $\lim_{n \to \infty} n^3 \left(\int_0^{\pi}\cosh(2\cos(x)) dx - \pi\sum_{k=0}^n \frac{1}{(k!)^2} \right)$

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

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A simple calculation shows that

$$\int_{0}^{\pi} \cosh(2\cos x) \, dx = \pi \sum_{k=0}^{\infty} \frac{1}{(k!)^2}.$$

Thus it remains to find the limit

$$\lim_{n\to\infty} \pi n^3 \sum_{k=n+1}^{\infty} \frac{1}{(k!)^2} .$$

But since

$$\sum_{k=n+1}^{\infty} \frac{1}{(k!)^2} \leq \frac{1}{(n!)^2} \sum_{k=n+1}^{\infty} \frac{1}{k^2} = O\left( \frac{1}{(n!)^2} \right),$$

it follows that the limit goes to zero.

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$$\int_0^{\pi} \cosh(2\cos x)dx=\sum_{k=0}^{\infty}\dfrac {2^{2k}}{(2k)!}\int_0^\pi \cos^{2k}xdx=\sum_{k=0}^{\infty}\dfrac{\pi \cdot2^{2k}}{(2k)!}\prod_{j=1}^{\infty}\dfrac{2j-1}{2j}=\pi \sum_{k=0}^{\infty}\dfrac{1}{(k!)^2}$$ 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.

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