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

The given series is a geometric series with the common ratio ( r = \frac{1}{4} ). It is conditionally convergent because its terms alternate in sign and the absolute value of its terms approaches zero as ( n ) approaches infinity.

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Answer 2

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

#(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#
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Answer 3

The geometric series converges absolutely, with

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

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#
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Answer 4

It converges absolutely.

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!

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Answer 5

The series is conditionally convergent.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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