Question

Use the properties of geometric series to find the sum of the given series. For what value of the variable does the series converge to this sum? 1+z/6+z^2/36+z^3/216+⋯ Enclose numerators, and denominators in parentheses. For example, (a−b)/(1+n). The sum of the given series is:

Answer 1

`|z| < 6` for
`f(z) =(6)/(6 - z)`

Explanation:

Given

`1+(z)/(6)+(z^2)/(36)+(z^3)/(216)+.....`

Required

The value at which the series converges

Calculate r

`r = (z)/(6)/1`

`r = (z)/(6)`

For a series to converge:

`|r| < 1`

This gives:

`|(z)/(6)| < 1`

`(1)/(6)|z| < 1`

Multiply both sides by 6

`6 * (1)/(6)|z| < 1 * 6`

`|z| < 6`

This is calculated using the sum to infinity of a gp.

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

Where
`a=1`

So:

`S = (1)/(1 - (z)/(6))`

Take LCM

`S = (1)/((6 - z)/(6))`

Rewrite as:

`S = 1 * {(6)/(6 - z)}`

`S =(6)/(6 - z)`

So, the function converges at:

`|z| < 6` for
`f(z) =(6)/(6 - z)`