Question

How do you evaluate `\sum _ { n = 1} ^ { \infty } \frac { ( - 3) ^ { n - 1} } { 7^ { n } }`?

Answer 1

`sum_(n=1)^(oo) (-3)^(n-1)/7^n = 1/10`

Explanation:

We want to evaluate the series summation:

`sum_(n=1)^(oo) (-3)^(n-1)/7^n`

Let us denote the `n^(th)` term in the series by `u_n`, then:

`u_n = (-3)^(n-1)/7^n`
`\ \ \ \ = (-3)/(-3)(-3)^(n-1)/7^n`
`\ \ \ \ = 1/(-3)(-3)^n/7^n`
`\ \ \ \ = -1/3(-3/7)^n`

So the first few sequence terms of the above series (starting `n=1`) are:

`-1/3(-3/7), -1/3(-3/7)^2, -1/3(-3/7)^3, -1/3(-3/7)^4, ...`

or:

`1/7, 1/7(-3/7), 1/7(-3/7)^2, 1/7(-3/7)^3...`

These terms form an Geometric Progression (GP) with:

`a=1/7` and `r=-3/7`

and using the standard GP formula `S_oo = a/(1-r)` we can calculate the summation;

`sum_(n=1)^(oo) (-3)^(n-1)/7^n = (1/7)/(1-(-3/7))`
`" " = (1/7)/(10/7)`
`" " = 1/10`

Answer 2

`sum_(n=1)^oo (-3)^(n-1)/7^n = 1/10`

Explanation:

`sum_(n=1)^oo (-3)^(n-1)/7^n = sum_(n=1)^oo 1/7(-3)^(n-1)/7^(n-1) = 1/7sum_(n=1)^oo (-3/7)^(n-1)`

If you change the summation index with `m=n-1`, then we have:

`sum_(n=1)^oo (-3)^(n-1)/7^n = 1/7 sum_(m=0)^oo (-3/7)^m`

this is the sum of a geometric progression, and we know that for `abs(x) < 1`:

`sum_(m=0)^oo x^m = 1/(1-x)`

so:

`1/7sum_(m=0)^oo (-3/7)^m = 1/7(1/(1+3/7))= (1/(7+3))=1/10`