How do i show that $x^n$ converges on $|x|=1$ only if $x=1$?

Question

The radius of convergence of $x^n$ is $1$.

So the convergence of it is inconclusive when $|x|=1$, but how do i prove that $x$ is divergent when $|x|=1$ but $x\neq 1$, not using gamma function?

Answer 1

first of all when we talk about radius of convergence we talk about series not sequences, you should say that the radius of convergence of this series is one$$\sum_{n=0}^{\infty}x^n$$ and now you want to stuyd the case when $|x|=1$ namely when $x=1$ and when $x=-1$. I'll assume that you know the divergent test which states that if $\sum_{n=0}^{\infty}a_n$ converges we must have $\lim_{n\to \infty}a_n=0$. in the case $x=1$ you will have $a_n=1$ which does not go to zero as $n$ go to infinity. similarly, when $x=-1$. which implies that the series diverges when $|x|=1$ and hence the interval of convergence is $(-1,1)$.

Edit: if we are in the complex case and $|x|=1$ then you will have $a_n=x^n$ does not converge to zero(otherwise we will have the following $1=|x|=|x|^n=|x^n|\to 0$ which is a contradiction) which implies that the series $\sum_{n=0}^{\infty}a_n$ does not converge.

Answer 2

When $x = 1$, you have $\sum_{k=0}^n x^k = n + 1$. What is happening as $n\to \infty$?

When $x = -1$, $\sum_{k=0}^n x^k$ "flip-flops." What does that say about its convergence?

You ask about the complex case. In that case, I'd tell you that $x^n\not\rightarrow 0$. A series whose terms fail to converge to zero cannot converge.

Answer 3

Suppose that the limit exists, and is equal to $w$. It is clear that $w\ne 0$. In order to make it clear that we are working in the complex plane, we use $z$ instead of $x$.

Since $\lim_{n\to\infty} z^n=\lim_{n\to\infty} z^{n+1}=x\lim_{n\to\infty}z^n$, we have $w=zw$, so $z=1$.

Answer 4

$x=1\cdot e^{i\varphi}$, hence for $\varphi\neq0$ $x^n$ is not a Cauchy sequence. (I assume, that $x$ is complex, as suggested in question).