How would I test for convergence at the following interval - $\sum\limits_{n=1}^\infty \frac{x^n}{\sqrt n}$ So I have the problem. 
$$ \sum _{ n-1 }^{ \infty  }{ \frac { { x }^{ n } }{ \sqrt { n }  }  } \quad $$
By the Ration test.
$$ \lim _{ n\rightarrow \infty  }{ \left| \frac { { a }_{ n+1 } }{ { a }_{ n } }  \right|  } \\ =\lim _{ n\rightarrow \infty  }{ \left| \frac { x\sqrt { n }  }{ \sqrt { n+1 }  }  \right|  } \\ =\left| x \right| \lim _{ n\rightarrow \infty  }{ \left| \frac { \sqrt { n }  }{ \sqrt { n+1 }  }  \right|  } \\ =\left| x \right| \lim _{ n\rightarrow \infty  }{ \left| \sqrt { \frac { n }{ n+1 }  }  \right|  } \\ =\left| x \right| \sqrt { \lim _{ n\rightarrow \infty  }{ \left| \frac { n }{ n+1 }  \right|  }  } \\ =\left| x \right| \sqrt { \lim _{ n\rightarrow \infty  }{ \left| \frac { \frac { n }{ n }  }{ \frac { n }{ n } +\frac { 1 }{ n }  }  \right|  }  } \\ =\left| x \right| \sqrt { 1 } \\ =\left| x \right| \quad \\ =-1<x<1$$
I am not seeing how to show  $$ [-1,1) $$
 A: You have established the convergence on $]-1,1[$, to see that the given serie converges at $x=-1$, use the alterning series test. The sequence $\left(\frac{1}{\sqrt{n}}\right)$ is decreasing toward $0$. Furthermore, if you want to see that the given serie does not converge at $x=1$, use an integral test for convergence. 
A: Plug $x=1$ into the series. You get the $\displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$. Clearly, this diverges by using a $p$-series argument with $p \leq 1$.  Also, it can be shown to diverge with the integral test.
If $x=-1$, then $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}$ will converge conditionally.
To see this, the absolute series will diverge (for the same reasons as above), but, $\displaystyle \lim_{n\rightarrow \infty} \frac{1}{\sqrt{n}} = 0$.
Also, $\displaystyle \frac{1}{\sqrt{n}}$ clearly defines a monotonically decreasing sequence for all $n \geq 1$. 
So the series converges conditionally by the alternating series test for $x=-1$. Therefore, the interval of convergence is $[-1,1)$.
