Question

Find the sum to infinity of the following arithmetico - geometric sequence:

`1, 2/4, 3/16, 4/64, ...`

Answer 1

Let S = `1 + 2/4 + 3/16 + 4/64 + ...`

i.e., S = `1 + 2/4 + 3/4^2 + 4/4^3 + ...`   ...(1)

∴ `1/4"S" = 1/4 + 2/4^2 + 3/4^3 + 4/4^4 + ...`  ...(2)

Subtracting (2) from (1), we get,

`3/4"S" = 1 + (1/4 + 1/4^2 + 1/4^3 + ......)`   ...(3)

The series in the bracket, i.e., `1/4 + 1/4^2 + 1/4^3 + ...` is a geometric series with a = `1/4`, r = `1/4`.

Since |r| = `|1/4| = 1/4 < 1`, the sum to infinity of this G.P. exists and

`1/4 + 1/4^2 + 1/4^3 + ...`

= `"a"/(1 - "r")`

= `((1/4))/(1 - (1/4))`

= `1/3`

∴ from (3), `3/4"S" = 1 + 1/3 = 4/3`

∴ S = `16/9`.

Alternative Method :

The given sequence can be written as:

` 1 xx 1, 2 xx 1/4, 3 xx 1/4^2, 4 xx 1/4^3 ...`

This is arithmetico-geometric progression with

a = 1, d = 1, r = `1/4`, where |r| = `1/4 < 1`

∴ sum to infinity of this A.G.P. is given by

S = `"a"/(1 - "r") + "dr"/(1 - "r")^2`

= `1/(1 - (1/4)) + (1. 1/4)/(1 - 1/4)^2`

= `4/3 + 1/4 xx 16/9`

= `4/3 + 4/9 = 16/9`