Question

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

Answer 1

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)`