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

Question

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

Evelyn Agard · Accepted Answer

#R=1# 2. #R=1/2#

According to the Ratio Test, we should let #L=lim_(n->oo)|a_(n+1)/a_n|#. Three scenarios can occur when working with Power Series. 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|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. Next, we have #|x|# We know if #|x|<1,# we have absolute convergence (and hence convergence), so #R=1.# Reduce complexity. #a_(n+1)=(2^n(2)(x-1)^n(x-1))/(n+1)# Division is multiplication by the reciprocal, so let's take the limit. #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|

Isaiah Allen · Answer

undefined The radius of convergence of a power series can be determined using the ratio test. The ratio test states that if the limit as n approaches infinity of the absolute value of the ratio of the (n+1)th term to the nth term is less than 1, then the series converges absolutely. The radius of convergence (R) is found by taking the limit as n approaches infinity of the absolute value of the ratio of consecutive terms, then applying the properties of limits. Mathematically, R is equal to the limit as n approaches infinity of the absolute value of (a_(n+1) / a_n).

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