Question

How do you find the sum of the infinite geometric series `Sigma 4(1/4)^n` from n=0 to `oo`?

Answer 1

`Sigma_0^oo 4(1/4)^n=5 1/3`

Explanation:

The first term of the series `Sigma_0^oo 4(1/4)^n` is

`4(1/4)^0=4xx1=4` and as each successive term `1/4` times its immediately preceding term common ratio is `1/4`.

and let the sum of the series `S_oo` be `4+1+1/4+1/16+1/64+...........`

As `S_oo=4+1+1/4+1/16+1/64+...........`, we have

`4S_oo=16+4+1+1/4+1/16+1/64+...........`

Now subtracting first from second we get

`3S_oo=16` and `S_oo=16/3=5 1/3`

Note - In a geometric series with `a` as first term and common ratio as `r`, `S_oo=a/(1-r)`, hence sum of series is `4/(1-1/4)=4/(3/4)=4xx4/3=16/3=5 1/3`.