Question

Consider the infinite geometric series (∞/Σ/n=1) * -4(1/3)^{n-1}.

a. Write the first 4 terms of the series.
b. Does the series converge or diverge?
c. If the series has a sum, find the sum.

Answer 1

The first four terms of the series probably refer to the first four partial sums.


`\displaystyle\sum_(n=1)^1-4\left(\frac13\right)^(n-1)=-4\left(\frac13\right)^(1-1)=-4`

`\displaystyle\sum_(n=1)^2-4\left(\frac13\right)^(n-1)=-\frac{16}3`

`\displaystyle\sum_(n=1)^3-4\left(\frac13\right)^(n-1)=-\frac{52}9`

`\displaystyle\sum_(n=1)^4-4\left(\frac13\right)^(n-1)=-(160)/(27)`

which you can compute either by adding one term at a time, or using the well-known formula,


`\displaystyle\sum_(n=1)^kar^(n-1)=a(1-r^(k+1))/(1-r)`

(I can provide a link to a derivation I gave in a nearly identical question in the comments)

The series converges as
`k\to\infty` if and only if
`|r|<1`, which is certainly the case here, since
`-1<\frac13<1`.

Extrapolating from the formula above, the sum of the convergent series is


`\displaystyle\sum_(n=1)^\infty ar^(n-1)=\frac a{1-r}`

so the sum for this series is
`(-4)/(1-\frac13)=-6`.