a. The series is
[tex]\displaystyle\sum_{n=1}^\infty-4\left(\frac13\right)^{n-1}=-4-\frac43-\frac4{3^2}-\frac4{3^3}-\cdots[/tex]
(first four terms are listed)
b. The series converges because this is a geometric series with [tex]r=\dfrac13<1[/tex].
c. Let [tex]S_N[/tex] be the [tex]N[/tex]-th partial sum of the series:
[tex]S_N=\displaystyle\sum_{n=1}^N-4\left(\frac13\right)^{n-1}[/tex]
[tex]S_N=-4-\dfrac43-\dfrac4{3^2}-\cdots-\dfrac4{3^{N-1}}[/tex]
Multiplying both sides by [tex]\dfrac13[/tex] gives
[tex]\dfrac13S_N=-\dfrac43-\dfrac4{3^2}-\dfrac4{3^3}-\cdots-\dfrac4{3^N}[/tex]
Subtracting this from [tex]S_N[/tex] gives
[tex]S_N-\dfrac13S_N=\dfrac23S_N=-4+\dfrac4{3^N}[/tex]
[tex]\implies S_N=-6+\dfrac6{3^N}[/tex]
As [tex]N[/tex] gets larger and larger [tex](N\to\infty)[/tex] the rational term converges to 0 and we're left with
[tex]\displaystyle\lim_{N\to\infty}S_N=\sum_{n=1}^\infty-4\left(\frac13\right)^{n-1}=-6[/tex]
