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...

3 Answers
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 + 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!

Apr 23, 2018

"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

Apr 23, 2018

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