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?
 A: This is an absolutely convergent series, therefore the series has to converge regardless of the minus signs.
A: The series can be written as the sum of positive terms and negative terms.
$$\sum_{i=1}^{\infty}(-3/4)^{i+1} = \sum_{k=0}^{\infty}(3/4)^{2k}-\sum_{k=0}^{\infty}(3/4)^{2k+1} = a_n - b_n$$
Both $a_n$ and $b_n$ being geometric series are convergent.
Hence, their sum should also be a convergent series. Proof of convergence of sum of two convergent series.
A: 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}
$
A: The Leibniz criterion  could be mentioned.
The series is convergent.
https://en.m.wikipedia.org/wiki/Alternating_series_test
