we have the series
[tex]\sum ^{\infty}_{n\mathop=1}-4(\frac{1}{3})^{(n-1)}[/tex]
Part a
Write the first four terms of the series.
First-term
For n=1
substitute
[tex]\begin{gathered} \sum ^{\infty}_{n\mathop{=}1}-4(\frac{1}{3})^{(1-1)} \\ \sum ^{\infty}_{n\mathop{=}1}-4(\frac{1}{3})^0=-4 \end{gathered}[/tex]
Second term
For n=2
[tex]\sum ^{\infty}_{n\mathop{=}1}(-4)-4(\frac{1}{3})^{(2-1)}=-4-\frac{4}{3}=-\frac{16}{3}[/tex]
Third term
For n=3
[tex]\sum ^{\infty}_{n\mathop{=}1}-\frac{16}{3}-4(\frac{1}{3})^{(3-1)}=-\frac{16}{3}-\frac{4}{9}=-\frac{52}{9}[/tex]
Fourth term
For n=4
[tex]\sum ^{\infty}_{n\mathop{=}1}-\frac{52}{9}-4(\frac{1}{3})^{(4-1)}=-\frac{52}{9}-\frac{4}{27}=-\frac{160}{27}[/tex]
therefore
the first four terms of the series are
-4,-16/3,-52/9,-160/27
Part B
Does the series diverge or converge?
Remember that
if −1In this problem
the common ratio r is equal to 1/3
so
r< 1
that means
The series converges
Part C
If the series has a sum, find the sum
In this problem
the series converges, and it has a sum.
the sum is equal to -6
