The convergence or divergence of a series depends on the sum of the series and its common ratio. The series converges because its sum is b
The series is given as:
[tex]\sum\limits^{\infty}_{n=1} 2 (\frac{2}{3})^n[/tex]
A geometric series is given as:
[tex]T_n = ar^{n-1}[/tex]
So, we rewrite the series as follows:
[tex]\sum\limits^{\infty}_{n=1} 2 (\frac{2}{3})^n = \sum\limits^{\infty}_{n=1} 2 (\frac{2}{3})^n[/tex]
Apply law of indices
[tex]\sum\limits^{\infty}_{n=1} 2 (\frac{2}{3})^n = \sum\limits^{\infty}_{n=1} 2 (\frac{2}{3})^{n-1} \times (\frac{2}{3})[/tex]
Rewrite
[tex]\sum\limits^{\infty}_{n=1} 2 (\frac{2}{3})^n = \sum\limits^{\infty}_{n=1} 2 \times (\frac{2}{3})(\frac{2}{3})^{n-1}[/tex]
[tex]\sum\limits^{\infty}_{n=1} 2 (\frac{2}{3})^n = \sum\limits^{\infty}_{n=1} \frac{4}{3}(\frac{2}{3})^{n-1}[/tex]
So, we have:
[tex]T_n = \frac{4}{3}(\frac{2}{3})^{n-1}[/tex]
Compare the above series to: [tex]T_n = ar^{n-1}[/tex]
[tex]a = \frac{4}{3}[/tex]
[tex]r = \frac{2}{3}[/tex]
The sum of the series to infinity is:
[tex]S_{\infty} = \frac{a}{1 - r}[/tex]
So:
[tex]S_{\infty} = \frac{4/3}{1 - 2/3}[/tex]
[tex]S_{\infty} = \frac{4/3}{1/3}[/tex]
[tex]S_{\infty} = 4[/tex]
i.e. [tex]\sum\limits^{\infty}_{n=1} 2 (\frac{2}{3})^n = 4[/tex]
Also:
The common ratio (r)
[tex]r = \frac{2}{3}[/tex]
[tex]r = \frac{2}{3}[/tex] [tex]< 1[/tex]
When common ratio is less than 1, then such series is convergent.
Hence, (b) is true because it has a sum of 4
Read more about series at:
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