Question

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 `(-1)^n` or `(-1)^(n+1)` or `(-1)^(n-1)`??

The exponents seem to be decreasing, i.e. `a^1`, `a^0`, `a^-1`, `a^-2`...

Answer 1

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 + 1/4 + 1/16 + ...`

This is a geometric series of common ratio `1/4`. Thus it converges. Since both `|a_n|` converges `a_n` converges absolutely.

Hopefully this helps!

Answer 2

`"It is a simple geometric series and it converges absolutely with"` `"sum "= 16/5 = 3.2."`

Explanation:

`(1+a+a^2+a^3+...)(1-a) = 1" , provided that |a|<1"`
`=> 1+a+a^2+a^3+...= 1/(1-a)`
`"Take "a = -1/4", then we have"`
`=> 1-1/4+1/16-1/64+... = 1/(1+1/4) = 1/(5/4) = 4/5`
`"Now our series is four times as much as the first term is 4."`
`"So our series"`
`4-1+1/4-1/16+... = 4*4/5 = 16/5 = 3.2`

Answer 3

The geometric series converges absolutely, with

`sum_(n=0)^ooa_n=16/5, sum_(n=0)^oo|a_n|=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)^ooa(r)^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`:

`-1/4=-1/4`

`(1/4)/(-1)=-1/4`

`(-1/16)/(1/4)=-1/16*4=-1/4`

`(1/64)/(-1/16)=1/64*-16=-1/4`

Thus, this series is geometric, with the common ratio `r=-1/4`, and the first term `a=4.`

We can write the series as

`sum_(n=0)^oo4(-1/4)^n`

Recall that a geometric series `sum_(n=0)^ooa(r)^n` converges to `a/(1-r)` if `|r|<1`. So, if it converges, we can also find its exact value.

Here, `|r|=|-1/4|=1/4<1`, so the series converges:

`sum_(n=0)^oo4(-1/4)^n=4/(1-(-1/4))=4/(5/4)=4*4/5=16/5`

Now, let's determine if it converges absolutely.

`a_n=4(-1/4)^n`

Strip out the alternating negative term:

`a_n=4(-1)^n(1/4)^n`

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

`|a_n|=4(1/4)^n`

Thus,

`sum_(n=0)^oo|a_n|=sum_(n=0)^oo4(1/4)^n`

We see `|r|=1/4<1`, so we still have convergence:

`sum_(n=0)^oo4(1/4)^n=4/(1-1/4)=4/(3/4)=4*4/3=16/3`

The series converges absolutely, with

`sum_(n=0)^ooa_n=16/5, sum_(n=0)^oo|a_n|=16/3`