What is the interval of convergence of #sum ( (x-2) ^n ) / ( n^2 + 1 ) #?

1 Answer
Feb 28, 2016

Answer:

The radius of convergence #R# of a power series #sum_(n=0)^(oo) a_n x^n# is given by #R = 1/(lim "sup"_(n->oo)root(n)(abs(a_n)))#.

Here, #sum_(n=0)^oo ( (x-2) ^n ) / ( n^2 + 1 )# converges #AA x in [1, 3].#

Explanation:

First , we have to write #sum_(n=0)^oo ( (x-2) ^n ) / ( n^2 + 1 )# in the form of #sum_(n=0)^(oo) a_n x^n#. So we will do a change of variable #y = x-2 iff x = 2 + y.#

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

Therefore, #a_n = 1 / ( n^2 + 1 ) = |a_n|.#

Now, let's compute #R# :

#lim "sup"_(n->oo)root(n)(abs(a_n)) = lim "sup"_(n->oo)1/root(n)( n^2 + 1 ).#

#"Yet, " root(n)( n^2 + 1 ) = exp(ln(root(n)( n^2 + 1 ))) = exp(1/nln( n^2 + 1 ))) = e^(ln( n^2 + 1 )/n).#

#"And " lim_(n->oo)ln( n^2 + 1 )/n = lim_(n->oo) (1/(n^2 + 1) * 2n) / 1 = lim_(n->oo) (2n)/(n^2 + 1) = lim_(n->oo) 2/(2n) = 0 " by Bernouilli l'Hôpital's rule".#

#"Thus, " lim_(n->oo) root(n)( n^2 + 1 ) = lim_(n->oo) e^(ln( n^2 + 1 )/n) = e^0 = 1.#

#=> R = lim "sup"_(n->oo)1/root(n)( n^2 + 1 )#

#= lim_(n->oo)1/root(n)( n^2 + 1 ) = 1/1 = 1.#

Therefore, #sum_(n=0)^oo ( y^n ) / ( n^2 + 1 )# converges #AA y in ]-1, 1[#

#iff sum_(n=0)^oo ( (x-2) ^n ) / ( n^2 + 1 )# converges #AA x in ]1, 3[.#

Now, we should check the endpoints of the interval.

  • If #x = 3# :

#sum_(n=0)^oo ( 1 ^n ) / ( n^2 + 1 ) = sum_(n=0)^oo 1 / ( n^2 + 1 )# converges because #n^2 + 1 >= n^2AA n > 0, n in NN iff 1/(n^2 + 1) <= 1/n^2 AA n > 0, n in NN# by the comparison test.

  • If #x = 1# :

#sum_(n=0)^oo ( (-1) ^n ) / ( n^2 + 1 )# converges because it converges absolutely (see the case #x = 3# above).

#"Therefore, " sum_(n=0)^oo ( (x-2) ^n ) / ( n^2 + 1 )# converges #AA x in [1, 3].#