Ytukyg

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?

This is an absolutely convergent series, therefore the series has to converge regardless of the minus signs.

If $|x| < 1$ then
$displaystylesum_{n=0}^{infty} x^n
=frac1{1-x}$.

Therefore

$begin{align}\
sum_{n=1}^{infty} left(frac{-3}{4}right)^{n+1}
&=left(frac{-3}{4}right)^2sum_{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}{4cdot 7}\
&=frac{9}{28}\
end{align}
$

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.

The Leibniz criterion could be mentioned.

The series is convergent.

https://en.m.wikipedia.org/wiki/Alternating_series_test