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$$\lim_{x\to0} \frac{\cosh x\cosh 2x\cosh 3x \cdots \cosh nx-1}{x^2}$$

How would you tackle this problem? Thanks.

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up vote 5 down vote accepted

You want to calculate $\displaystyle\lim_{x\rightarrow 0}\frac{\displaystyle\prod_{k=1}^n\cosh kx - 1}{x^2}$. Let's start by noticing that $\displaystyle\prod_{k=1}^n\cosh kx - 1 = \displaystyle\prod_{k=1}^n\cosh kx - \displaystyle\prod_{k=1}^{n-1}\cosh kx + \displaystyle\prod_{k=1}^{n-1}\cosh kx -1 = (\displaystyle\prod_{k=1}^{n-1}\cosh kx)(\cosh nx -1) + \displaystyle\prod_{k=1}^{n-1}\cosh kx -1$.

Now let's call $u_n = \displaystyle\lim_{x\rightarrow 0}\frac{\displaystyle\prod_{k=1}^n\cosh kx - 1}{x^2}$. From the previous equality, we have $u_n = u_{n-1} + \frac{n^2}{2}$. (Note: it is straightforward to prove by recurrence that $\forall n\in\mathbb{N},u_n\in\mathbb{R}$), and because $u_1 = \frac{1}{2}$, we have $u_n = \frac{1}{2}\displaystyle\sum_{k=1}^n k^2 = \displaystyle\frac{n(n+1)(2n+1)}{12}$

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Note that $$ \cosh \alpha=1+\frac{\alpha^2}{2}+o(\alpha^2) $$ for $\alpha\to0$. So for $x\to 0$ we have $$ \prod\limits_{k=1}^n \cosh kx= \prod\limits_{k=1}^n \left(1+\frac{k^2 x^2}{2}+o(k^2 x^2)\right)= \prod\limits_{k=1}^n \left(1+\frac{k^2 x^2}{2}+o(x^2)\right)= 1+\sum\limits_{k=1}^n\frac{1}{2}k^2 x^2+o(x^2) $$ Hence $$ \lim\limits_{x\to 0}\frac{\prod\limits_{k=1}^n\cosh kx - 1}{x^2}= \lim\limits_{x\to 0}\frac{\frac{1}{2}\sum\limits_{k=1}^n k^2 x^2+o(x^2)}{x^2}= \frac{1}{2}\sum\limits_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{12} $$

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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

$$ \lim_{x\to 0} \frac{\cosh x \cosh 2x \cdots \cosh nx - 1}{x^2} = \lim_{x\to 0} \frac{\frac{n(n+1)(2n+1)}{12}x^2+O(x^4)}{x^2} =\frac{n(n+1)(2n+1)}{12}\,. $$

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Duplicate of this answer – Norbert Sep 6 '12 at 9:50

I should want to leave my answer in comment box but my partial answer is too long. I try the following , I know there are a lot of calculations but I think what I've done can help us to find out the answer.

$\displaystyle \frac{\cosh x\cosh 2x\cosh 3x \cdots \cosh nx-1}{x^2}=\frac{(e^x+e^{-x})(e^{2x}+e^{-2x})\cdot \ldots \cdot(e^{nx}+e^{-nx})-2^n}{2^n \cdot x^2}=$ $$\large \frac{\left(\frac{e^{2x}+1}{e^x}\right)\cdot \left(\frac{e^{4x}+1}{e^{2x}}\right)\cdot \ldots\cdot \left(\frac{e^{2nx}+1}{e^{nx}}\right)-2^{n}}{2^n \cdot x^2}.$$ I want to write this expression beautifully. $e^x =a$. So we have:

$$\large\frac{\frac{(a^2+1)(a^4+1)\ldots(a^{2n}+1)}{a^{\frac{n(n+1)}{2}}}-2^n}{2^n \cdot x^2}=\frac{(a^2+1)(a^4+1)\ldots(a^{2n}+1)-2^n \cdot a^{\frac{n(n+1)}{2}}}{2^{n}\cdot x^{2} \cdot a^{\frac{n(n+1)}{2}}}. $$

$$=\large \frac{(a^2-1)(a^2+1)(a^4+1)\ldots(a^{2n}+1)-2^n \cdot a^{\frac{n(n+1)}{2}}\cdot(a^2-1)}{2^{n}\cdot x^{2} \cdot a^{\frac{n(n+1)}{2}}\cdot (a^2-1)}$$

$$\large=\frac{(a^{4n}-1)-2^{n}\cdot a^{\frac{n(n+1)}{2}}\cdot(a^2-1)}{2^{n}\cdot x^{2} \cdot a^{\frac{n(n+1)}{2}}\cdot (a^2-1)}$$

$$\large=\frac{e^{4nx}-1-2^{n} \cdot e^{\frac{n(n+1)}{2}\cdot x +2x}+2^{n}\cdot e^{\frac{n(n+1)}{2}\cdot x}}{2^{n} \cdot e^{\frac{n(n+1)}{2}\cdot x +2x} \cdot x^{2} -2^{n}\cdot e^{\frac{n(n+1)}{2}\cdot x} \cdot x^{2}}.$$

$$\large \lim_{x \rightarrow 0}{\frac{e^{4nx}-1-2^{n} \cdot e^{\frac{n(n+1)}{2}\cdot x +2x}+2^{n}\cdot e^{\frac{n(n+1)}{2}\cdot x}}{2^{n} \cdot e^{\frac{n(n+1)}{2}\cdot x +2x} \cdot x^{2} -2^{n}\cdot e^{\frac{n(n+1)}{2}\cdot x} \cdot x^{2}}}=$$ and I think this limit can be calculate with L'Hospital.

Is it ok?

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