Are You Sure that's not -1?

It's not $-1$ , so let's make it $-1$

$s_t = \displaystyle{\lim_{n \to \infty}} \sum_{r=1}^{n} \dfrac{2^r}{t^{2^r}+1}$ = $s_t = \sum_{r=1}^{\infty} \dfrac{2^r}{t^{2^r}+1}$ = $s_t = \sum_{r=1}^{\infty} \dfrac{(2^r)(t^{2^r}-1)}{t^{2^{r+1}}-1}$ = $s_t = \sum_{r=1}^{\infty} \dfrac{(2^r)(t^{2^r}+1-2)}{t^{2^{r+1}}-1}$ = $s_t = \sum_{r=1}^{\infty} \dfrac{(2^r)}{t^{2^r}-1}-\dfrac{2^{r+1}}{t^{2^{r+1}}-1}$ = $\dfrac{2}{t^2-1}$ . Hence, $\dfrac{s_7}{s_{17}} = 6$

Let us denote as $s_t(n)$ the sum of the first $n$ terms of $s_t$ , i.e.:

$\large \displaystyle s_t(n) = \sum_{r=1}^n \frac{2^r}{t^{2^r} + 1}$

Looking at the first terms:

$\large \displaystyle s_t(1) = \frac{2}{t^2+1}$

$\large \displaystyle s_t(2) = \frac{2}{t^2+1} + \frac{4}{t^4+1} = \frac{2(t^4 + 2t^2 + 3)}{t^6 + t^4 + t^2 + 1}$

$\large \displaystyle s_t(3) = \frac{2}{t^2+1} + \frac{4}{t^4+1} + \frac{8}{t^8+1} = \frac{2(t^{12} + 2t^{10} + 3t^8 + 4t^6 + 5t^4 + 6t^2 + 7)}{t^{14} + t^{12} + t^8 + t^6 + t^4 + t^2 + 1}$

As a general formula:

$\large \displaystyle s_t(n) = \frac{2 \left [ t^{2^{n+1} - 4} + 2t^{2^{n+1} - 6} + 3t^{2^{n+1} - 8} + ... + (2^n-3)t^4 + (2^n-2)t^2 + (2^n-1) \right ]}{t^{2^{n+1}-2} + t^{2^{n+1}-4} + ... + t^2 + 1}$

Let us define:

$\large \displaystyle K = t^{2^{n+1} - 4} + 2t^{2^{n+1} - 6} + 3t^{2^{n+1} - 8} + ... + (2^n-3)t^4 + (2^n-2)t^2 + (2^n-1)$

$\large \displaystyle Kt^2 = t^{2^{n+1} - 2} + 2t^{2^{n+1} - 4} + 3t^{2^{n+1} - 6} + ... + (2^n-3)t^2 + (2^n-2)t^2 + (2^n-1)$

Subtracting them:

$\large \displaystyle K(t^2-1) = t^{2^{n+1} -2 } + t^{2^{n+1} - 4} + ... + t^2 + 1 - 2^n$

$\large \displaystyle K(t^2-1) = \frac{t^{2^{n+1}}-1}{t^2-1} - 2^n$

$\color{#20A900} \boxed { \large \displaystyle K = \frac{t^{2^{n+1}}-1 - 2^n(t^2-1)}{(t^2-1)^2} }$ .

So:

$\large \displaystyle s_t(n) = \frac{2K}{t^{2^{n+1}-2} + t^{2^{n+1}-4} + ... t^2 + 1}$

$\large \displaystyle s_t(n) = \frac{ 2\frac{t^{2^{n+1}}-1 - 2^n(t^2-1)}{(t^2-1)^2} }{ \frac{t^{2^{n+1}}-1}{t^2-1} }$

$\color{#20A900} \boxed { \large \displaystyle s_t(n) = \frac{2}{t^2-1} \cdot \frac{t^{2^{n+1}}-1 - 2^n(t^2-1)}{t^{2^{n+1}}-1} }$

Then, $s_t$ will be the limit of $s_t(n)$ as $n$ goes to $\infty$ :

$\large \displaystyle s_t = \lim_{n \rightarrow \infty} s_t(n)$

$\large \displaystyle s_t = \frac{2}{t^2-1} \cdot \lim_{n \rightarrow \infty} \frac{t^{2^{n+1}}-1 - 2^n(t^2-1)}{t^{2^{n+1}}-1}$

It is easy to see that this limit is $1$ . Then:

$\color{#20A900} \boxed { \large \displaystyle s_t = \frac{2}{t^2-1} }$

Thus:

$\color{#3D99F6} \large \displaystyle s_7 = \frac{1}{24}, s_{17} = \frac{1}{144}, \boxed{\large \displaystyle \frac{s_7}{s_{17}} = 6}$

Write $x = 1/t$ so that the sum is given by $s_t = \sum_{r=1}^{\infty} \frac{2^r x^{2^r}}{1 + x^{2^r}}.$ Now log-differentiating the formula $\frac{1}{1-x^2} = \prod_{r=1}^{\infty}(1 + x^{2^r}), \qquad (|x| < 1)$ which easily follows by letting $n \to \infty$ to $1 - x^{2^n} = (1 - x^2)(1 + x^2)(1 + x^{2^2}) \cdots (1 + x^{2^{n-1}})$ , we obtain $\frac{2x}{1-x^2} = \sum_{r=1}^{\infty} \frac{2^r x^{2^r-1}}{1 + x^{2^r}},$ So it follows that $s_t = \frac{2x^2}{1-x^2} = \frac{2}{t^2 - 1}.$ From this, we easily find that $s_7 / s_{17} = 6$ .