Prove that if a series converges to $a$, then that series to the power $k$ converges to $a^k$ Let $a_n$ be a sequence in the real numbers. Prove that:
$a_n\rightarrow a \implies (a_n)^k \rightarrow a^k$   $\forall k \in N$ 
I think I need to do this by induction. The base case is simple. When $k=1$:
$(a_n)^k = (a_n)^1=a_n \rightarrow a =  a^1=a^k$.
Now assume $(a_n)^i \rightarrow a^i$. Then there exists $N>0$ such that if $n>N$, then $\mid (a_n)^i- a^i \mid < \epsilon$. I need to show that $(a_n)^{i+1} \rightarrow a^{i+1}$. So I think I need to manipulate $n$ and $N$, right? 
Any ideas on how to approach the problem? 
Thanks,
K
 A: When using induction, you are probably encouraged to use that if $a_n\to a$ and $b_n\to b$ then $a_nb_n\to ab$. Thus, assuming $a_n^{k-1}\to a^{k-1}$, you get $$\lim\limits_{n\to\infty} a_n^k=\lim\limits_{n\to\infty}a_n^{k-1}\lim\limits_{n\to\infty}a_n=a^{k-1}a=a^k$$ using the inductive hypothesis in the second equality.
A: The following is a fundamental property of continuous functions $\mathbb{R}\to \mathbb{R}$: 
If $f:\mathbb{R}\to \mathbb{R}$ is a continuous function, and if $(a_n)_{n\in\mathbb{N}}$ is a sequence in $\mathbb{R}$ with limit $a\in \mathbb{R}$, then $(f(a_n))_{n\in\mathbb{N}}$ is a sequence in $\mathbb{R}$ with limit $f(a)\in\mathbb{R}$.
The claim holds in a much more general context but as stated it is all that is relevant to your question. Can you see how to answer your question using this claim?
If you can, then the next step is to prove this claim. I encourage you to try as it is an elementary $\epsilon$-$\delta$ argument but if you're stuck, then I'm happy to help.
The converse of this claim is also true; as an exercise, it would be good to state and prove the converse precisely.
I hope this helps!
