Consider $\sum_{k=1}^{\infty}{1\over 3^k}\cdot{\sinh(a/3^k)+\sinh(2a/3^k)\over 2\cosh(a/3^k)+\cosh(2a/3^k)+1.5}=S_a$

Question

Consider

$$\sum_{k=1}^{\infty}{1\over 3^k}\cdot{\sinh(a/3^k)+\sinh(2a/3^k)\over 2\cosh(a/3^k)+\cosh(2a/3^k)+1.5}=S_a\tag1$$ $a\ge1$

How does one show that $$S_a={1\over 2}\cdot{e^a+1\over e^a-1}-{1\over a}?$$

Try to simplify

$x={a\over 3^k}$

$${\sinh(x)+\sinh(2x)\over 2\cosh(x)+\cosh(2x)+1.5}={\sinh(x)(1+2\cosh(x))\over 2\cosh(x)+2\cosh^2(x)+0.5}$$

$$={2\sinh(x)(1+2\cosh(x))\over 3\cosh^2(x)+4\cosh(x)-\sinh^2x}$$

I don't how to simplify down any more.

Edited:

$$\sum_{k=1}^{\infty}{1\over 3^k}\cdot{2\sinh(x)\over 1+2\cosh(x)}$$

$$\sum_{k=1}^{\infty}{1\over 3^k}\cdot{1+2e^{a/3^k}\over 1+2\cosh(a/3^k)}=S_a+{1\over2}$$

Answer 1

Detailing @JackD'Aurizio Answer, $$ \begin{align} \sum_{k=1}^{\infty}\,\frac{1}{3^k}\,\frac{2\sinh\left(a/{\small3^k}\right)}{1+2\cosh\left(a/{\small3^k}\right)} &= \frac{1}{2}\frac{e^a+1}{e^a-1}-\frac{1}{a} \quad\Rightarrow \\[3mm] \sum_{k=1}^{\infty}\,\frac{1}{3^k}\,\frac{4\sinh\left(a/{\small3^k}\right)}{1+2\cosh\left(a/{\small3^k}\right)} &= \coth(a/{\small2})-\frac{1}{a/{\small2}} \end{align} $$ The idea is that, instead of simplifying the series of LHS to get the expression on RHS,
Start by the expression on RHS and construct the series of LHS. $$ \begin{align} \color{red}{\coth(3x)} &= \frac{\cosh(3x)}{\sinh(3x)} = \frac{\cosh(x+2x)}{\sinh(x+2x)} = \frac{\cosh(x)\cosh(2x)+\sinh(x)\sinh(2x)}{\sinh(x)\cosh(2x)+\cosh(x)\sinh(2x)} \\[3mm] &= \frac{\cosh(x)\left[2\sinh^2(x)+1\right]+\sinh(x)\left[2\sinh(x)\cosh(x)\right]}{\sinh(x)\left[2\cosh^2(x)-1\right]+\cosh(x)\left[2\sinh(x)\cosh(x)\right]} \\[3mm] &= \frac{4\sinh^2(x)\cosh(x)+\cosh(x)}{4\cosh^2(x)\sinh(x)-\sinh(x)} = \frac{1}{3}\frac{12\sinh^2(x)\cosh(x)+3\cosh(x)}{4\cosh^2(x)\sinh(x)-\sinh(x)} \\[3mm] &= \frac{1}{3}\frac{8\sinh^2(x)\cosh(x)+2\cosh(x)\left[2\sinh^2(x)+1\right]+\cosh(x)}{2\sinh(x)\left[2\cosh^2(x)-1\right]+\sinh(x)} \\[3mm] &= \frac{1}{3}\frac{\sinh(x)\left[4\sinh(2x)\right]+\cosh(x)\left[1+2\cosh(2x)\right]}{\sinh(x)\left[1+2\cosh(2x)\right]} \\[3mm] &= \color{red}{\frac{1}{3}\frac{4\sinh(2x)}{1+2\cosh(2x)} + \frac{1}{3}\coth(x)} \end{align} $$
Thus, $$ \begin{align} \color{red}{\coth(a/{\small2})} &= \frac{1}{3^1}\,\frac{4\sinh\left(a/{\small3^1}\right)}{1+2\cosh\left(a/{\small3^1}\right)} + \frac{1}{3^1}\,\coth\left(\frac{a/{\small2}}{\small3^1}\right) \\[3mm] &= \frac{1}{3^1}\,\frac{4\sinh\left(a/{\small3^1}\right)}{1+2\cosh\left(a/{\small3^1}\right)} + \frac{1}{3^2}\,\frac{4\sinh\left(a/{\small3^2}\right)}{1+2\cosh\left(a/{\small3^2}\right)} + \frac{1}{3^2}\,\coth\left(\frac{a/{\small2}}{\small3^2}\right) \\[3mm] &= \,\cdots\, = \sum_{k=1}^{\infty}\,\frac{1}{3^k}\,\frac{4\sinh\left(a/{\small3^k}\right)}{1+2\cosh\left(a/{\small3^k}\right)} + \color{blue}{\lim_{k\rightarrow\infty}\left[\frac{1}{3^k}\,\coth\left(\frac{a/{\small2}}{\small3^k}\right)\right]} \quad\Rightarrow \\[3mm] \sum_{k=1}^{\infty}\,\frac{1}{3^k} &\, \frac{4\sinh\left(a/{\small3^k}\right)}{1+2\cosh\left(a/{\small3^k}\right)} = \color{red}{\coth(a/{\small2})} - \color{blue}{\frac{1}{a/{\small2}}} \end{align} $$ Where, $$ \lim_{k\rightarrow\infty}\left[\frac{1}{3^k}\,\coth\left(\frac{a/{\small2}}{\small3^k}\right)\right] = \frac{1}{a/{\small2}}\lim_{k\rightarrow\infty}\left[\frac{a/{\small2}}{3^k}\,\coth\left(\frac{a/{\small2}}{\small3^k}\right)\right] = \frac{1}{a/{\small2}} $$

Answer 2

Since you already have a conjectural $S_a$ that is related with $\coth\left(\frac{a}{2}\right)$, and the given series is screaming for creative telescoping, the most reasonable thing is to check how we may simplify $$ 3a\coth(3a/2)-a\coth(a/2) $$ through the (hyperbolic) cotangent triplication formulas. If that gives (essentially) the general term of the series, we are fine. And indeed, it does: $$ 3\coth(3x)-\coth(x) = \frac{8 \cosh(x)\sinh(x)}{1+2 \cosh(x)^2+2 \sinh(x)^2}.$$ At last, we just need to exploit the fact that $\lim_{x\to 0^+}x\coth(x)=1$.