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:

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MrRoyal MrRoyal · Answer 1 · 2021-03-26T09:12:57+01:00

Answer:

[tex]|z| < 6[/tex] for [tex]f(z) =\frac{6}{6 - z}[/tex]

Step-by-step explanation:

Given

[tex]1+\frac{z}{6}+\frac{z^2}{36}+\frac{z^3}{216}+.....[/tex]

Required

The value at which the series converges

Calculate r

[tex]r = \frac{z}{6}/1[/tex]

[tex]r = \frac{z}{6}[/tex]

For a series to converge:

[tex]|r| < 1[/tex]

This gives:

[tex]|\frac{z}{6}| < 1[/tex]

[tex]\frac{1}{6}|z| < 1[/tex]

Multiply both sides by 6

[tex]6 * \frac{1}{6}|z| < 1 * 6[/tex]

[tex]|z| < 6[/tex]

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

[tex]S = \frac{a}{1- r}[/tex]

Where [tex]a=1[/tex]

So:

[tex]S = \frac{1}{1 - \frac{z}{6}}[/tex]

Take LCM

[tex]S = \frac{1}{\frac{6 - z}{6}}[/tex]

Rewrite as:

[tex]S = 1 * {\frac{6}{6 - z}}[/tex]

[tex]S =\frac{6}{6 - z}[/tex]

So, the function converges at:

[tex]|z| < 6[/tex] for [tex]f(z) =\frac{6}{6 - z}[/tex]