Let [tex]r[/tex] be the common ratio between terms. Then
[tex]-2.88=4r\implies r=-0.72[/tex]
Now denote the [tex]n[/tex]th partial sum of the series by
[tex]S_n=4-2.88+2.0736-1.492992+\cdots+4(-0.72)^{n-1}+4(-0.72)^n[/tex]
Multiply both sides by [tex]-0.72[/tex], then subtract from the above to get
[tex]S_n-(-0.72S_n)=4-4(-0.72)^{n+1}[/tex]
[tex]1.72S_n=4(1-(-0.72)^{n+1})[/tex]
[tex]S_n=2.32558(1-(-0.72)^{n+1})[/tex]
As [tex]n\to\infty[/tex], you're left with
[tex]S_\infty=2.32558[/tex]
