Determine if the power series converges or diverges for each value of x?

1) #sum_(n=0)^(\infty )##(x^n)/(2^n)# for x = -1
2)#sum_(n=0)^(\infty )##((x-1)^n)/(3^n)# for x = 5

2 Answers
VNVDVI
Apr 6, 2018
  1. Converges by the Alternating Series Test 2. Diverges by Geometric Series Test

Explanation:

  1. Simply plug in #x=-1# into #a_n=x^n/2^n.#

This yields #sum_(n=0)^oo(-1)^n/2^n#

This is an alternating series in the form #a_n=(-1)^nb_n#.

Here, we see #b_n=1/2^n#.

#b_n>=b_(n+1)# for all #n;# the denominator is always increasing for larger #n# values causing the series to always be decreasing.

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

So, we have convergence by the Alternating Series Test.

(We could also rewrite as #sum_(n=0)^oo(-1/2)^n# and verify convergence by the fact that this is a geometric series in the form #sumar^n# with #|r|=|-1/2|<1#)

  1. Again, plug in #x=5,# yielding #sum_(n=0)^oo(5-1)^n/3^n=sum_(n=0)^oo4^n/3^n#

We can rewrite as

#sum_(n=0)^oo(4/3)^n#

This resembles a geometric series in the form #suma(r)^n# where #r=4/3.#

Well, recall that for a geometric series, we have convergence if #|r|<1# and divergence otherwise. Here, #|r|=4/3>1,# so we have divergence.

Hammer
Apr 6, 2018

The first series converges while the second diverges.

Explanation:

First case:

1) #sum_(n=0)^oo x^n/2^n# for #x=-1#.

By substituting #x# into the series, we have

#sum_(n=0)^oo (-1/2)^n#

We know that

#1+r+r^2+r^3+...=1/(1-r)# if and only if #|r|<1#.

In our case, #r=-1/2#. Since #|r| <1#, the power series does #color(red)("converge")#.

#:. sum_(n=0)^oo (-1/2)^n=1/(1+1/2) = 2/3#

Second case:

2)#sum_(n=0)^oo(x-1)^n/3^n#, for #x=5#.

#=> sum_(n=0)^oo (4/3)^n#

However, as said above, the geometric series only converges if #|r|<1#. Since #4/3# is #"bigger"# than 1, the series #color(red)("diverges")#.

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