Compute $\lim_{x\to0} \frac{\cosh x\cosh 2x\cosh 3x \cdots \cosh nx-1}{x^2}$ Compute:
$$\lim_{x\to0} \frac{\cosh x\cosh 2x\cosh 3x \cdots \cosh nx-1}{x^2}$$
How would you tackle this problem? Thanks.

By Taylor's theorem we have $\cosh x = 1 + \frac{x^2}2 + O(x^4)$. So ignoring the parts in the product which tend to $0$ faster than $x^2$ we get
$$ \cosh x \cosh 2x \cdots \cosh nx = \left(1+\frac{x^2}2\right)\left(1+\frac{(2x)^2}2\right)\cdots\left(1+\frac{(nx)^2}2\right)+O(x^4) $$
The product of polynomials can be easily calculated if we ignore the terms of more than quadratic order. We get
$$ \cosh x \cosh 2x \cdots \cosh nx = 1+\frac{x^2}2+\frac{(2x)^2}2+\cdots+\frac{(nx)^2}2+O(x^4) $$
The coefficient of $x^2$ is half the sum of the first $n$ square numbers which is $n(n+1)(2n+1)/12$. Hence we get
$$ \lim_{x\to 0} \frac{\cosh x \cosh 2x \cdots \cosh nx - 1}{x^2} 
= \lim_{x\to 0} \frac{\frac{n(n+1)(2n+1)}{12}x^2+O(x^4)}{x^2}
=\frac{n(n+1)(2n+1)}{12}\,. $$
