Convergence of the series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{x^n}$. 
Show that $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{x^n}$$ converges for every $x>1$.

let $a(x)$ be the sum of the series. does $a$ continious at $x=2$? differentiable?
I guess the first part is with leibniz but I am not sure about it.

Hint:
Use the geometric series:
$$a(x) = \frac1{x}\sum_{n=0}^\infty \frac{(-1)^n}{x^n} = \frac{1}{x\left(1+\frac1x\right)} = \frac{1}{x+1}$$
