Question

How do you calculate `sum_(n=1)^(∞)(3/2)^(1-2n)`?

How do you calculate `sum_(n=1)^(∞)(3/2)^(1-2n)`?

Answer 1

`sum_(n=1)^oo (3/2)^(1-2n) = 6/5`

Explanation:

Note that:

`sum_(n=1)^oo (3/2)^(1-2n) = sum_(n=1)^oo (2/3) (3/2)^(2-2n)`

`color(white)(sum_(n=1)^oo (3/2)^(1-2n)) = sum_(n=1)^oo (2/3) (2/3)^(2n-2)`

`color(white)(sum_(n=1)^oo (3/2)^(1-2n)) = sum_(n=1)^oo (2/3) (4/9)^(n-1)`

This is now in the standard form of a geometric series with initial term `a=2/3` and common ratio `4/9`.

Given any geometric series, the general term can be written:

`a_n = a r^(n-1)`

where `a` is the initial term and `r` the common ratio.

Then we find:

`(1-r) sum_(n=1)^N a r^(n-1) = sum_(n=1)^N a r^(n-1) - r sum_(n=1)^N a r^(n-1)`

`color(white)((1-r) sum_(n=1)^N a r^(n-1)) = sum_(n=1)^N a r^(n-1) - sum_(n=2)^(N+1) a r^(n-1)`

`color(white)((1-r) sum_(n=1)^N a r^(n-1)) = a+color(red)(cancel(color(black)(sum_(n=2)^N a r^(n-1)))) - color(red)(cancel(color(black)(sum_(n=2)^N a r^(n-1)))) - a r^N`

`color(white)((1-r) sum_(n=1)^N a r^(n-1)) = a - a r^N`

`color(white)((1-r) sum_(n=1)^N a r^(n-1)) = a(1 - r^N)`

Dividing both ends by `(1-r)` we find:

`sum_(n=1)^N a r^(n-1) = (a(1 - r^N))/(1-r)`

So if `abs(r) < 1` then:

`sum_(n=1)^oo a r^(n-1) = lim_(N->oo) sum_(n-1)^N a r^(n-1) = lim_(N->oo) (a(1-r^N))/(1-r) = a/(1-r)`

Applying this in our example with `a=2/3` and `r=4/9` we find:

`sum_(n=1)^oo (3/2)^(1-2n) = sum_(n=1)^oo (2/3) (4/9)^(n-1) = (2/3)/(1-4/9) = (2/3)/(5/9) = 6/5`