To prove x_n is summable under the given norm

Question

Let X be a normed space, and $x_n$ $∈$ $X$, n = 1,2,3,...If $X$ is a Banach space and $lim$ $sup$||$x_n$||$^1$$^/$$^n$ < 1, then $\sum_{n=1}^\infty x_n$ is summable.

I don't understand how to proceed with the given norm?

Answer 1

Fix $\alpha$ , $\limsup ||x_n||^{1/n} < \alpha < 1$. Then from the definition of $\limsup$ you can find an

$n_{0} \in \Bbb N$ such that $\forall n\geq n_{0}$ then $||x_n||^{1/n}<\alpha.$So you get $\sum_{n=n_{0}}^{\infty} ||x_n|| \leq \sum_{n=n_{0}}^{\infty} \alpha ^n<\infty.$

Now since $X$ is a Banach space and the series $\sum_{n=n_{0}}^{\infty} ||x_n||$ converges it follows that the series

$\sum_{n \in \Bbb N}^{\infty} x_n$ is to summabale to some point $x\in X$.