I need to show whether the following infinite series converges, and if it does, to which value. Here, $t$ is a constant: $$\sum_{x=1}^{\infty}{e^{tx}(\frac{1}{2})^{x}}$$ This is my solution: $$\sum_{x=1}^{\infty}{e^{tx}(\frac{1}{2})^{x}} = \sum_{x=1}^{\infty}{e^{t}e^{x}(\frac{1}{2})^{x}} = e^{t}\sum_{x=1}^{\infty}{(\frac{e}{2})^{x}}$$ $$|\frac{e}{2}| > 1$$ Therefore it diverges.
The solution, however, is that it converges to $\frac{e^t}{2-e^t}$. What is my mistake here?
As pointed out by @charlus, ${e^{tx} \neq e^t e^x}$. Instead, you have:
$${=\sum_{x=1}^{\infty}\left(e^t\right)^{x}0.5^x}$$
$${=\sum_{x=1}^{\infty}\left(\frac{e^t}{2}\right)^x}$$
This is a geometric series. Provided ${\left(\frac{e^t}{2}\right) < 1}$ (as pointed out by @charlus again, this implies ${t < \ln(2)}$), we have
$${=\sum_{x=0}^{\infty}\left(\frac{e^t}{2}\right)^{x}-1=\frac{1}{1-\frac{e^t}{2}}-1}$$
(I started the index from $0$ and just subtracted $1$ at the end so we can use ${\sum_{n=0}^{\infty}r^n = \frac{1}{1-r}}$). Simplifying you get
$${=\frac{2}{2-e^t}-\frac{2-e^t}{2-e^t}=\frac{e^t}{2-e^t}}$$