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

Question

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

Luke Alvarez · Accepted Answer

Converges by the Alternating Series Test 2. Diverges by Geometric Series Test 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# Therefore, the Alternating Series Test indicates convergence. (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#) We can reword 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.

William Abdalla · Answer

The first series converges while the second diverges.

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")#.

Ezekiel Alexander · Answer

undefined To determine if a power series converges or diverges for a specific value of $ x $, you typically need to analyze its convergence properties using tests such as the ratio test, root test, or comparison test. Without specifying the power series, it's challenging to provide a definitive answer. Could you please provide the power series in question?

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

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