Prove that $\sinh{2u}+2\sinh{4u}+3\sinh{6u}+...+n\sinh{2nu}=\frac{n\sinh{(2n+2)u-(n+1)\sinh{2nu}}}{4\sinh^2{u}}$

Question

Prove that $$\sinh{2u}+2\sinh{4u}+3\sinh{6u}+...+n\sinh{2nu}=\frac{n\sinh{(2n+2)u-(n+1)\sinh{2nu}}}{4\sinh^2{u}}$$

My attempt at a solution:

Let $$S=\sum_{r=1}^{n}\cosh{2ru}$$ then $$\frac{dS}{du}=\sum_{r=1}^{n}2r\sinh{2ru}\Rightarrow\sum_{r=1}^{n}{r\sinh{2ru}}=\frac{1}{2}\frac{dS}{du}$$ To evaluate $S$, I used $\cosh{2ru}=\frac{1}{2}{(e^{2ru}+e^{-2ru})}$, from which $$S=\frac{1}{2}\left\lbrace\sum_{r=1}^{n}e^{2ru}+\sum_{r=1}^n{e^{-2ru}}\right\rbrace =\frac{1}{2}\left\lbrace\frac{e^{2u}((e^{2u})^n-1)}{e^{2u}-1}+\frac{e^{-2u}(1-(e^{-2u})^n)}{1-e^{-2u}}\right\rbrace,$$ using the formula for the sum of the first $n$ terms of a geometric progression.

After some algebra and cleaning up, I managed to obtain $$S=\frac{\sinh(2n+1)u}{2\sinh{u}}-\frac{1}{2}$$ and so $$\frac{dS}{du}=\frac{1}{2}\left[\frac{(\sinh{u})(2n+1)\cosh{(2n+1)u}-(\sinh{(2n+1)u})\cosh{u}}{\sinh^2{u}}\right]$$ but I struggle to spot the relevant hyperbolic identities (if needed) in order to proceed to the given result.

Just curious, but is there an alternative method to reach the desired result?

Answer 1

$$\sinh(x\pm y) = \sinh x \cosh y \pm \cosh x \sinh y$$ Hence \begin{align}&\quad(\sinh u)(2n+1)\cosh((2n+1)u)-\sinh((2n+1)u)\cosh u \\~\\&= (n+1)(\sinh u\cosh ((2n+1)u) - \sinh((2n+1)u)\cosh u) \\&\;\;+n(\sinh u\cosh ((2n+1)u) + \sinh((2n+1)u)\cosh u) \\~\\&=(n+1)\sinh(u-(2n+1)u)+n\sinh(u+(2n+1)u) \\~\\&=(n+1)\sinh(-2nu)+n\sinh((2n+2)u) \\~\\&=n\sinh((2n+2)u)-(n+1)\sinh(2nu) \end{align}

Answer 2

Proof by induction:

1. I leave it up to to show that the relation is valid for $n=1$, it should be trivial.

2. Induction step:

$$S_n=\frac{n\sinh{(2n+2)u-(n+1)\sinh{2nu}}}{4\sinh^2{u}}$$

$$S_{n+1}=S_n+(n+1)\sinh2(n+1)u=\frac{n\sinh{(2n+2)u-(n+1)\sinh{2nu}}}{4\sinh^2{u}}+(n+1)\sinh(2n+2)u$$

$$S_{n+1}=\frac{n\sinh{(2n+2)u-(n+1)\sinh{2nu}}+4(n+1)\sinh^2u\sinh(2n+2)u}{4\sinh^2{u}}$$

$$S_{n+1}=\frac{n\sinh{(2n+2)u-(n+1)\sinh{2nu}}+2(n+1)(\cosh2u-1)\sinh(2n+2)u}{4\sinh^2{u}}$$

$$S_{n+1}=\frac{(-n-2)\sinh{(2n+2)u-(n+1)\sinh{2nu}}+2(n+1)\cosh2u\sinh(2n+2)u}{4\sinh^2{u}}$$

$$S_{n+1}=\frac{-(n+2)\sinh{(2n+2)u-(n+1)\sinh{2nu}}+(n+1)(\sinh(2n+4)u+\sinh2nu)}{4\sinh^2{u}}$$

$$S_{n+1}=\frac{(n+1)\sinh{(2n+4)u-(n+2)\sinh{(2n+2)u}}}{4\sinh^2{u}}$$

This completes the induction step.

Formulas used:

$$2\sinh^2 x=\cosh2x-1$$

$$2\sinh x \cosh y = \sinh(x+y)+\sinh(x-y)$$