What is the interval of convergence of `sum_1^oo ((-1)^(-n)*x^(-n))/sqrtn`?

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

The series `suma_n` converges if `lim_(nrarroo)abs(a_(n+1)/a_n)<1` through the ratio test. So, we will find this ratio and solve for the values of `x` that make `L<1`.

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

Simplifying:

`L=lim_(nrarroo)abs((-1)^-1x^-1sqrt(n/(n+1)))`

The `-1` can be ignored since we're working within the absolute value. The `x` term can be pulled from the limit because the limit only depends on the changing value of `n`.

`L=abs(1/x)lim_(nrarroo)abssqrt(n/(n+1))`

The limit approaches `1` as `nrarroo`. Thus

`L=abs(1/x)`

So the series converges when

`abs(1/x)<1`

This can be split up into

`-1<1/x<1`

Splitting into two inequalities, we see that `-1<1/x` can be solved through

`0<1/x+1=(1+x)/x>0`

Which is true on `x<-1` and `x>0`.

The other inequality `1/x<1` gives

`1/x-1>0=>(1-x)/x>0`

Which is true on `0 < x < 1`.

The intersection of the two solutions we found is `0 < x < 1`.

Before we call this our interval of convergence, plug the endpoints `x=0` and `x=1` into the original series and check to see if the series diverges or converges.

At `x=0`:

`sum_(n=1)^oo((-1)^-n0^-n)/sqrtn=sum_(n=1)^oo0`

This converges because it's always `0`. Thus `0` will be included in the interval of convergence.

At `x=1`:

`sum_(n=1)^oo((-1)^-n1^-n)/sqrtn`

`1^-n=1` for all values of `n` and `(-1)^-n=(-1)^n` as well, so this is just the alternating series

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

Which converges through the alternating series test. Since both `x=0` and `x=1` cause the series to converge, the interval of convergence is

`0lt=xlt=1`