How do you find the interval of convergence `Sigma x^n/(6sqrtn)` from `n=[1,oo)`?

The series `suma_n` converges when `L=lim_(nrarroo)abs(a_(n+1)/a_n)<1`. We can find `L` then find the values of `x` that satisfy `L<1`.

`L=lim_(nrarroo)abs(a_(n+1)/a_n)=lim_(nrarroo)abs((x^(n+1)/(6sqrt(n+1)))/(x^n/(6sqrtn)))=lim_(nrarroo)abs(x^(n+1)/x^n((6sqrtn)/(6sqrt(n+1)))`

We can simplify and move the `x` terms out of the limit, since the limit depends only on how `n` changes.

`L=absxlim_(nrarroo)abssqrt(n/(n+1))`

The limit approaches `1`:

`L=absx`

The series converges when `L<1`:

`absx<1`

`-1ltxlt1`

Before we say we're done, we have to check the endpoints `x=-1` and `x=1` by plugging them into the original series and seeing if the series does in fact converge. If the series converges for the endpoint, the endpoint will be included in the interval.

At `x=1`, the series in question is:

`sum_(n=1)^oo1^n/(6sqrtn)=1/6sum_(n=1)^oo1/sqrtn`

This is divergent through the p-series test, since `p=1/2`. Thus `x=1` is not included in the interval of convergence.

At `x=-1`, we have

`1/6sum_(n=1)^oo(-1)^n/sqrtn`

Through the alternating series test, we see that this does converge, so `x=-1` will be included in the interval of convergence.

The interval is then:

`-1lt=xlt1`