Explanation:[tex]\begin{gathered} Given:\text{ } \\ \text{a = -5} \\ r\text{ = 4/5} \end{gathered}[/tex]
The formula relating the sum of infinite series of a geometric sequence:
[tex]\sum_{n\mathop{=}1}^{\infty}a_n\text{ = }\frac{a}{1-r}[/tex][tex]\begin{gathered} The\text{ sum of infinite series of the Geometric sequence = }\frac{a}{1-r} \\ substitute\text{ the values:} \\ S_n\text{ = }\frac{-5}{1-\frac{4}{5}} \\ S_n\text{ = }\frac{-5}{\frac{5-4}{5}} \\ S_n\text{ = }\frac{-5}{\frac{1}{5}} \\ S_n\text{ = -5 }\times\text{ }\frac{5}{1}\text{ = -25} \\ \\ S_n\text{ = S}_{\infty}\text{ = -25} \\ \\ S_{\infty_}\text{ = -25} \end{gathered}[/tex]For the series to converge:
[tex]\begin{gathered} For\text{ the series to converge:} \\ \lim_{n\to\infty}S_n\text{ =}\sum_{n\mathop{=}1}^{\infty}a_n\text{ = actual value} \\ \\ For\text{ the series to diverge:} \\ \lim_{n\to\infty}S_n\text{ = }\pm\infty \\ \\ Also\text{ if \mid r\mid < 1, it converges} \\ if\text{ \mid r\mid \ge 1, it diverges} \end{gathered}[/tex][tex]\begin{gathered} Since\text{ }\sum_{n\mathop{=}1}^{\infty}a_n\text{ = -25 \lparen an actual value\rparen} \\ then\text{ }\operatorname{\lim}_{n\to\infty}S_n\text{ converges} \\ \\ |r|\text{ < 1, hence it converges} \end{gathered}[/tex]