Question

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

Answer 1

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].`