Convergence of $\sum_{n=1}^{\infty}\left(\frac{\sin(2\alpha^n)}{\alpha^n}-2\right)$ Determine values of $\alpha \in \left[\frac{1}{2} ; \frac{3}{2}\right]=:I$ s.t. the following series converges:
$$\sum_{n=1}^{\infty}\left(\frac{\sin(2\alpha^n)}{\alpha^n}-2\right)$$
Am I doing it wrong or this series fails the limit test for all of the $\alpha\in I$? In case I got the wrong, how to handle this series?
Edit:  I handled $2\alpha^n$ as $(2α)^n$ when evaluating the limit and so it goes to $0$ when $α<1$ and for the other values of α the result is $−2$.

Let $a_n = \frac{\sin(2\alpha^n)}{\alpha^n}-2$ be the general term for your series.
If $\alpha = 1$, then the $a_n$ are constant and equal to $\sin(2) - 2 \neq 0$, so the series diverges.
If $\alpha > 1$, then $\lim_{n\to\infty} a_n = -2$.
Since it is not $0$, the series does not converge.
Now, suppose $\alpha<1$, and write $a_n = 2\left(\frac{\sin(2\alpha^n)}{2\alpha^n}-1\right)$$.
We have have
$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$$
so that for $x \neq 0$
$$\frac{\sin(x)}x - 1 = - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \dots$$
In particular, for $0<x<1$, we have
$$- \frac{x^2}{3!} < \frac{\sin(x)}x - 1 < 0 \implies
\left|\frac{\sin(x)}x - 1\right| < \frac{x^2}{3!}.$$
It hence follows that
\begin{align}
\frac12 S
&=
\sum_{n=1}^{\infty}\,\left(\frac{\sin(2\alpha^n)}{2\alpha^n}-1\right)
\\&\leqslant
\sum_{n=1}^{\infty}\,\left|\frac{\sin(2\alpha^n)}{2\alpha^n}-1\right|
\\&\leqslant
\sum_{n=1}^{k}\,\left|\frac{\sin(2\alpha^n)}{2\alpha^n}-1\right|
+
\sum_{n=k+1}^{\infty}\,\frac{{(2\alpha^n)}^2}{3!}
\\&=
\sum_{n=1}^{k}\,\left|\frac{\sin(2\alpha^n)}{2\alpha^n}-1\right|
+
\frac23\,\sum_{n=k+1}^{\infty}\,{\left(\alpha^2\right)}^n < \infty,
\end{align}
where $k$ is such that $n>k \implies 2\alpha^n < 1$, and the last step follows from the fact that $|\alpha^2|<1$ so that the geometric series converges.
In particular, $S$ converges absolutely and hence converges.

While the bound on $\sin(x)/x - 1$ is in fact always valid, I suppose it's easier to convince yourself purely from the power series expansion that it holds on the interval $[0,1]$.
