Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

2 Answers 2

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.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.