Question

Consider the infinite geometric series ∑∞ n=1 -4(1/3)^n-1

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

Answer 1

a. The series is

`\displaystyle\sum_(n=1)^\infty-4\left(\frac13\right)^(n-1)=-4-\frac43-\frac4{3^2}-\frac4{3^3}-\cdots`

(first four terms are listed)

b. The series converges because this is a geometric series with
`r=\frac13<1`.

c. Let
`S_N` be the
`N`-th partial sum of the series:

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

`S_N=-4-\frac43-\frac4{3^2}-\cdots-\frac4{3^(N-1)}`

Multiplying both sides by
`\frac13` gives

`\frac13S_N=-\frac43-\frac4{3^2}-\frac4{3^3}-\cdots-\frac4{3^N}`

Subtracting this from
`S_N` gives

`S_N-\frac13S_N=\frac23S_N=-4+\frac4{3^N}`

`\implies S_N=-6+\frac6{3^N}`

As
`N` gets larger and larger
`(N\to\infty)` the rational term converges to 0 and we're left with

`\displaystyle\lim_(N\to\infty)S_N=\sum_(n=1)^\infty-4\left(\frac13\right)^(n-1)=-6`