Does $\sum _{n=1}^{\infty \:\:}\left(\frac{-3}{4}\right)^{n+1}$ converge? This is a geometric series, where $a=(3/4)^2$  and $r=-3/4$, meaning that it converges to $9/28$ right? The answer in my textbook says that it diverges, whereas Symbolab says it converges absolutely. So does it converge or diverge?

If $|x| < 1$ then
$\displaystyle\sum_{n=0}^{\infty} x^n
=\frac1{1-x}$.
Therefore
$\begin{align}\\
\sum_{n=1}^{\infty} \left(\frac{-3}{4}\right)^{n+1}
&=\left(\frac{-3}{4}\right)^2\sum_{n=1}^{\infty} \left(\frac{-3}{4}\right)^{n-1}\\
&=\frac{9}{16}\sum_{n=0}^{\infty} \left(\frac{-3}{4}\right)^{n}\\
&=\frac{9}{16}\frac1{1-\left(\frac{-3}{4}\right)}\\
&=\frac{9}{16}\frac{4}{4+3}\\
&=\frac{9}{4\cdot 7}\\
&=\frac{9}{28}\\
\end{align}
$
