Could you please help with this taylor series?

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1 Answer
Mar 12, 2018

The series:

#ln(4x^2+1) = sum_(n=0)^oo (-1)^n 2^(2n+2) x^(2n+2)/(n+1)#

is convergent for:

#-1/2 < x < 1/2#

Explanation:

The Taylor series for #f(x) = ln(1+x)# is:

#ln(x+1) = sum_(n=0)^oo (-1)^n x^(n+1)/(n+1)#

using the ratio test:

#lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) abs( ( x^(n+2)/(n+2))/(x^(n+1)/(n+1)))#

#lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) (n+1)/(n+2) abs(x^(n+2)/x^(n+1))#

#lim_(n->oo) abs(a_(n+1)/a_n) = absx lim_(n->oo) (n+1)/(n+2) #

#lim_(n->oo) abs(a_(n+1)/a_n) = absx#

we can see that the radius of convergence of the series is #R=1#.

The series:

#ln(4x^2+1) = sum_(n=0)^oo (-1)^n (4x^2)^(n+1)/(n+1)#

#ln(4x^2+1) = sum_(n=0)^oo (-1)^n 2^(2n+2) x^(2n+2)/(n+1)#

is therefore convergent for:

#abs (4x^2) < 1#

that is for:

#-1/2 < x < 1/2#