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

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!

#(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#

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#