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

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Answer 1 · 2017-12-21T12:34:49

Sequence converges.

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

Let's look at a few terms of this sequence.

#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#

We are asked whether or not the sequence converges.

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)#

Hence, in our example:

#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#