# Is the series indicated absolutely convergent, conditionally convergent, or divergent? rarr\4-1+1/4-1/16+1/64...

## Use the appropriate test. I get the feeling this is an Alternating Series, but I'm not sure if it's ${\left(- 1\right)}^{n}$ or ${\left(- 1\right)}^{n + 1}$ or ${\left(- 1\right)}^{n - 1}$?? The exponents seem to be decreasing, i.e. ${a}^{1}$, ${a}^{0}$, ${a}^{-} 1$, ${a}^{-} 2$...

Apr 23, 2018

It converges absolutely.

#### Explanation:

Use the test for absolute convergence. If we take the absolute value of the terms we get the series

$4 + 1 + \frac{1}{4} + \frac{1}{16} + \ldots$

This is a geometric series of common ratio $\frac{1}{4}$. Thus it converges. Since both $| {a}_{n} |$ converges ${a}_{n}$ converges absolutely.

Hopefully this helps!

Apr 23, 2018

$\text{It is a simple geometric series and it converges absolutely with}$ $\text{sum "= 16/5 = 3.2.}$

#### Explanation:

$\left(1 + a + {a}^{2} + {a}^{3} + \ldots\right) \left(1 - a\right) = 1 \text{ , provided that |a|<1}$
$\implies 1 + a + {a}^{2} + {a}^{3} + \ldots = \frac{1}{1 - a}$
$\text{Take "a = -1/4", then we have}$
$\implies 1 - \frac{1}{4} + \frac{1}{16} - \frac{1}{64} + \ldots = \frac{1}{1 + \frac{1}{4}} = \frac{1}{\frac{5}{4}} = \frac{4}{5}$
$\text{Now our series is four times as much as the first term is 4.}$
$\text{So our series}$
$4 - 1 + \frac{1}{4} - \frac{1}{16} + \ldots = 4 \cdot \frac{4}{5} = \frac{16}{5} = 3.2$

Apr 23, 2018

The geometric series converges absolutely, with

${\sum}_{n = 0}^{\infty} {a}_{n} = \frac{16}{5} , {\sum}_{n = 0}^{\infty} | {a}_{n} | = \frac{16}{3}$

#### Explanation:

This series is definitely an alternating series; however, it also looks geometric.

If we can determine the common ratio shared by all terms, the series will be in the form

${\sum}_{n = 0}^{\infty} a {\left(r\right)}^{n}$

Where $a$ is the first term and $r$ is the common ratio.

We'll need to find the summation using the above format.

Divide each term by the term before it to determine the common ratio $r$:

$- \frac{1}{4} = - \frac{1}{4}$

$\frac{\frac{1}{4}}{- 1} = - \frac{1}{4}$

$\frac{- \frac{1}{16}}{\frac{1}{4}} = - \frac{1}{16} \cdot 4 = - \frac{1}{4}$

$\frac{\frac{1}{64}}{- \frac{1}{16}} = \frac{1}{64} \cdot - 16 = - \frac{1}{4}$

Thus, this series is geometric, with the common ratio $r = - \frac{1}{4}$, and the first term $a = 4.$

We can write the series as

${\sum}_{n = 0}^{\infty} 4 {\left(- \frac{1}{4}\right)}^{n}$

Recall that a geometric series ${\sum}_{n = 0}^{\infty} a {\left(r\right)}^{n}$ converges to $\frac{a}{1 - r}$ if $| r | < 1$. So, if it converges, we can also find its exact value.

Here, $| r | = | - \frac{1}{4} | = \frac{1}{4} < 1$, so the series converges:

${\sum}_{n = 0}^{\infty} 4 {\left(- \frac{1}{4}\right)}^{n} = \frac{4}{1 - \left(- \frac{1}{4}\right)} = \frac{4}{\frac{5}{4}} = 4 \cdot \frac{4}{5} = \frac{16}{5}$

Now, let's determine if it converges absolutely.

${a}_{n} = 4 {\left(- \frac{1}{4}\right)}^{n}$

Strip out the alternating negative term:

${a}_{n} = 4 {\left(- 1\right)}^{n} {\left(\frac{1}{4}\right)}^{n}$

Take the absolute value, causing the alternating negative term to vanish:

$| {a}_{n} | = 4 {\left(\frac{1}{4}\right)}^{n}$

Thus,

${\sum}_{n = 0}^{\infty} | {a}_{n} | = {\sum}_{n = 0}^{\infty} 4 {\left(\frac{1}{4}\right)}^{n}$

We see $| r | = \frac{1}{4} < 1$, so we still have convergence:

${\sum}_{n = 0}^{\infty} 4 {\left(\frac{1}{4}\right)}^{n} = \frac{4}{1 - \frac{1}{4}} = \frac{4}{\frac{3}{4}} = 4 \cdot \frac{4}{3} = \frac{16}{3}$

The series converges absolutely, with

${\sum}_{n = 0}^{\infty} {a}_{n} = \frac{16}{5} , {\sum}_{n = 0}^{\infty} | {a}_{n} | = \frac{16}{3}$