What is the interval of convergence of #sum_1^oo (3x-2)^(n)/(1+n+n^(x) #?

1 Answer
Dec 6, 2016

The series converges for #1/3<=x<1#

Explanation:

We can use the ratio test and see for what values of #x#:

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

Calculate:
#L= lim_(n->oo) |(3x-2)^(n+1)/(1+n+1+(n+1)^x)(1+n+n^x)/(3x-2)^n|#

#L= lim_(n->oo) |(3x-2)^(n+1)/(3x-2)^n||(1+n+n^x)/(2+n+(n+1)^x)|#

#L= |(3x-2)|lim_(n->oo) |(1+n+n^x)/(2+n+(n+1)^x)|= |(3x-2)|#

So the series is certainly convergent for:

#|(3x-2)| < 1#

that is for:

#1/3 < x < 1#

And certainly divergent for #x < 1/3" and "x>1#

On the boundaries, where #L=1# the test is in inconclusive, we must analyse the two cases:

For #x=1/3# the series becomes:

#sum_1^oo (-1)^n/(1+n+n^(1/3))#

If we write this as:

#-sum_1^oo (-1)^(n+1)/(1+n+n^(1/3))#

We can see that for every #n# we have that:

#(-1)^(n+1)/(1+n+n^(1/3)) < (-1)^(n+1)/n#

and as the second series is the alternate armonic series that is convergent, so is the first.

For #x=1# the series becomes:

#sum_1^oo 1/(1+2n)#

We can us the integral test to show that this series is not convergent:

#lim_(x->oo) int_1^x dt/(1+2t) = -ln3 + lim_(x->oo) 1/2ln(1+2x) = +oo#

In conclusion the series converges for #x in [1/3,1)#