What is the radius of convergence by using the ratio test?

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

1 Answer
VNVDVI
Apr 6, 2018
  1. #R=1# 2. #R=1/2#

Explanation:

The Ratio Test tells us that we let

#L=lim_(n->oo)|a_(n+1)/a_n|#.

When dealing with Power Series, one of three cases can arise.

a. The limit tends to #oo,# meaning we only have a radius of convergence of #R=0#

b. The limit tends to zero, meaning #R=oo#

c. The most frequent case, we have absolute convergence (and hence convergence) for #|x-a|<R#.

So, let's apply the test.

  1. #a_n=n^3x^n#

#a_(n+1)=(n+1)^3x^(n+1)#

#a_(n+1)=x^n(x)(n+1)^3#

#lim_(n->oo)|(cancel(x^n)(x)(n+1)^3)/(n^3cancelx^n)|#

We can factor out the absolute value of #x,# as it's not the variable of our limit.

#|x|lim_(n->oo)(n+1)^3/n^3#

We drop the absolute value bars on the limit as we know these terms are positive as #n# grows.

We then have

#|x|#

We know if #|x|<1,# we have absolute convergence (and hence convergence), so #R=1.#

  1. #a_n=(2^n(x-1)^n)/n#
    #a_(n+1)=(2^(n+1)(x-1)^(n+1))/(n+1)#

Simplify.

#a_(n+1)=(2^n(2)(x-1)^n(x-1))/(n+1)#

Take the limit. Division is multiplication by the reciprocal.

#lim_(n->oo)|(cancel(2^n)(2)cancel((x-1)^n)(x-1))/(n+1)*n/(cancel(2^n)cancel((x-1)^n))|#

We can factor out #2|x-1|.#

#2|x-1|lim_(n->oo)n/(n+1)#

So, if #2|x-1|<1,# we have absolute convergence. We divide both sides by two to get this in the form #|x-1|<R#

#|x-1|<1/2#

Thus,

#R=1/2#

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