How do you find the interval of convergence `Sigma (3^n+4^n)x^n` from `n=[0,oo)`?

You can use the ratio test, evaluating:

`abs(a_(n+1)/a_n) = abs(((3^(n+1)+4^(n+1))x^(n+1))/((3^n+4^n)x^n))`

`abs(a_(n+1)/a_n) = ((3^(n+1)+4^(n+1)))/((3^n+4^n))abs(x)`

`abs(a_(n+1)/a_n) = 4^(n+1)/4^n((3/4)^(n+1)+1)/((3/4)^n+1)abs(x)`

`abs(a_(n+1)/a_n) = 4((3/4)^(n+1)+1)/((3/4)^n+1)abs(x)`

and passing to the limit for `n->oo`:

`lim_(n->oo) abs(a_(n+1)/a_n) = 4abs(x)`

So the series is absolutely convergent for `abs(x) < 1/4` and divergent for `abs(x) > 1/4`.

As usual, the case where `lim_(n->oo) abs(a_(n+1)/a_n) = 1` must be examined separately:

1. For `x = 1/4` the series becomes:

`sum_(n=0)^oo (3^n+4^n)/4^n = sum_(n=0)^oo[ (3/4)^n+1]`

which clearly is divergent as `s_n > n`, where `s_n` is the `n`-th partial sum.

1. For `x = -1/4` the series becomes:

`sum_(n=0)^oo (-1)^n(3^n+4^n)/4^n = sum_(n=0)^oo(-1)^n[ (3/4)^n+1]`

We can see that this series is not convergent as:

`abs(s_(n+1) = s_n) = (3/4)^(n+1) +1 > 1`

so that the series does not satisfy Cauchy's convergence test.

`sum_(n=0)^oo(3^n+4^n)x^n=sum_(n=0)^oo(3x)^n+sum_(n=0)^oo(4x)^n`

`sum_(n=0)^m(3x)^n = ((3x)^(m+1)-1)/(3x-1)`

`sum_(n=0)^m(4x)^n = ((4x)^(m+1)-1)/(4x-1)`

Defining `I_3=abs(3x) < 1` and `I_4 = abs(4x)<1` we have

`I_4 sub I_3`

Now regarding `sum_(n=0)^m(4x)^n` we now that if `x in I_4` we have

`lim_(m->oo)sum_(n=0)^m(4x)^n=-1/(4x-1)=1/(1-4x)` and

`lim_(m->oo)sum_(n=0)^m(3x)^n=-1/(3x-1)=1/(1-3x)`

Finally we have that

`sum_(n=0)^oo(3^n+4^n)x^n=1/(1-3x)+1/(1-4x)` for `x in I_4`