Find the sum of the series. 1 + z 3 + z2 9 + z3 27 + $ correct: your answer is correct. for what values of the variable does the series converge to this sum? (enter your answer using interval notation.)

[tex]1+\dfrac z3+\dfrac{z^2}9+\dfrac{z^3}{27}+\cdots=\displaystyle\sum_{n=0}^\infty\left(\frac z3\right)^n[/tex]

which converges for [tex]\left|\dfrac z3\right|<1[/tex] or [tex]|z|<3[/tex] to [tex]\dfrac1{1-\frac z3}=\dfrac3{3-z}[/tex].

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MrRoyal MrRoyal · Answer 2 · 2022-03-22T14:37:11+01:00

The sum to infinity of the series is: [tex]S_{\infty} = \frac{3}{3 - z}[/tex], and it converges from [tex]|z| < 3[/tex] to [tex]\frac{3}{3 - z}[/tex]

The series is given as:

[tex]1 + \frac z3 + \frac{z^2}9 + \frac{z^3}{27} + ........[/tex]

Notice that the series has no end.

This means that, we calculate the sum of the series to infinity.

The series is a geometric series, and it has the following parameters

Initial value, a = 1

Common ratio, r = z/3

The sum to infinity of a geometric series is:

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

So, we have:

[tex]S_{\infty} = \frac{1}{1 - z/3}[/tex]

Multiply by 3/3

[tex]S_{\infty} = \frac{3}{3 - z}[/tex]

Set the denominator to 0

[tex]3 - z = 0[/tex]

Solve for z

[tex]z= 3[/tex]

In the actual sense, z can be positive or negative.

So, we have:

[tex]|z| = 3[/tex]

Since the series converges, it means that the common ratio (r) is less than 1.

i.e. r < 1

So, we have:

[tex]|z| < 3[/tex]

This means that the series converges from [tex]|z| < 3[/tex] to [tex]\frac{3}{3 - z}[/tex]

Read more about geometric series at:

https://brainly.com/question/12006112