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Does #a_n=(-1/2)^n# sequence converge or diverge? How do you find its limit?

Answer 1

Ezra Adams

Sequence converges.

#a_n = (-1/2)^n#

Let's examine a few terms found in this list.

#a_1 = -1/2#

#a_2 = 1/4#

#a_3 = -1/8#

This is a geometric progresion (GP) with first term #a_1 =-1/2# and common ratio #(r) = -1/2#

The question of whether the sequence converges is posed to us.

Consider, #lim_(n->oo) a_n = lim_(n->oo) (-1/2)^n =0#

#:.# the sequence will converge.

Now let's consider the sum of the infinite series #sum_(n=1)^oo a_n#

The sum of an infinite GP where #absr<0# is given by #a_1/(1-r)#

Thus, in our illustration:

#sum_(n=1)^oo a_n = (-1/2)/(1-(-1/2)#

#= -1/(2(3/2))#

#=-1/3#

So, we can state that the sequence converges and the sum of the infinite sequence converges to #-1/3#

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Answer 2

Ezra Adams

The sequence ( a_n = (-1/2)^n ) converges. Its limit can be found by analyzing the behavior of the terms as ( n ) approaches infinity. Since the absolute value of ( (-1/2)^n ) approaches 0 as ( n ) becomes large, the sequence converges to 0.

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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.