Question

Which statement describes ?

The series diverges because it has a sum of 4.
The series converges because it has a sum of 4.
The series diverges because it does not have a sum.
The series converges because it does not have a sum.

Answer 1

Answer: B

The series converges because it has a sum of 4.

Step-by-step explanation:

Answer 2

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:

$\sum\limits^{\infty}_{n=1} 2 (\frac{2}{3})^n$

A geometric series is given as:

$T_n = ar^{n-1}$

So, we rewrite the series as follows:

$\sum\limits^{\infty}_{n=1} 2 (\frac{2}{3})^n = \sum\limits^{\infty}_{n=1} 2 (\frac{2}{3})^n$

Apply law of indices

$\sum\limits^{\infty}_{n=1} 2 (\frac{2}{3})^n = \sum\limits^{\infty}_{n=1} 2 (\frac{2}{3})^{n-1} \times (\frac{2}{3})$

Rewrite

$\sum\limits^{\infty}_{n=1} 2 (\frac{2}{3})^n = \sum\limits^{\infty}_{n=1} 2 \times (\frac{2}{3})(\frac{2}{3})^{n-1}$

$\sum\limits^{\infty}_{n=1} 2 (\frac{2}{3})^n = \sum\limits^{\infty}_{n=1} \frac{4}{3}(\frac{2}{3})^{n-1}$

So, we have:

$T_n = \frac{4}{3}(\frac{2}{3})^{n-1}$

Compare the above series to: $T_n = ar^{n-1}$

$a = \frac{4}{3}$

$r = \frac{2}{3}$

The sum of the series to infinity is:

$S_{\infty} = \frac{a}{1 - r}$

So:

$S_{\infty} = \frac{4/3}{1 - 2/3}$

$S_{\infty} = \frac{4/3}{1/3}$

$S_{\infty} = 4$

i.e. $\sum\limits^{\infty}_{n=1} 2 (\frac{2}{3})^n = 4$

Also:

The common ratio (r)

$r = \frac{2}{3}$

$r = \frac{2}{3}$ $< 1$

When common ratio is less than 1, then such series is convergent.

Hence, (b) is true because it has a sum of 4

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