If a sequence $\{S_n\}$ converges to a constant $s$, then the sequence $\{S_n^k\}$ converges to $s^k$ Let $\{s_n\}$ be a sequence in $\mathbb{R}$, and assume that $s_n \rightarrow s$. Prove that $s^k_n\rightarrow s^k$ for every $k \in\mathbb{N}$
Ok, so we need $|s^k_n - s^k| < \varepsilon$. I rewrote this as
$$|s_ns^{k-1}_n - ss^{k-1}|=|(s_n-s)(s^{k-1}_n + s^{k-1}) -s_ns^{k-1}+ss_n^{k-1}|$$
But this seems really messy. What should I use here: $|s_n - s| < \varepsilon?$
Help!
 A: Hint: $$|s_n^k-s^k|=|s_n-s|\cdot |s_n^{k-1}+s_n^{k-2}s+\dots +s_ns^{k-2}+s^{k-1}|\\\leq |s_n-s|\cdot k(|s|+|s_n-s|)^k$$
A: It suffices to show that the function $f(x) = x^k$ is continuous.  Now clearly the function $g(x) = x$ is continuous ($\delta = \epsilon$).  Now just prove that the product of continuous functions is continuous and then use induction.
A: Since $s_n \to s$,
there is an $n_1$
such that
$|s_n-s|
\le s_0
$
for $n  > n_1$,
where
$s_0 = \max(|s|, 1)$.
(The "max" is to take care
of the possibility that
$s = 0$.)
This implies that
$|s_n| \le |s|+s_0$.
Note that the purpose of this
is to get a bound on
$s_n$ that depends only
on $s$ and constants.
Then,
as in tohecz's answer,
for $n > n_1$,
$\begin{align}
|s_n^k-s^k|
&=|s_n-s|\big|\sum_{j=0}^{k-1} s_n^j s^{k-1-j}\big|\\
&\le|s_n-s|\sum_{j=0}^{k-1} \big|s_n^j s^{k-1-j}\big|\\
&\le|s_n-s|\sum_{j=0}^{k-1} \big|s_n^j\big| \big|s^{k-1-j}\big|\\
&\le|s_n-s|\sum_{j=0}^{k-1} \big|(|s|+s_0)^j\big| \big|s^{k-1-j}\big|\\
&\le|s_n-s|\sum_{j=0}^{k-1} (|s|+s_0)^{k-1}\\
&=|s_n-s|k (|s|+s_0)^{k-1}\\
&=b(s, k)|s_n-s|\\
\end{align}
$
where
$b(s, k)
=k (|s|+s_0)^{k-1}
$.
To make
$|s_n^k-s^k|
< \epsilon
$,
it is sufficient if
$b(s, k)|s_n-s|
< \epsilon
$,
or
$|s_n-s|
< \dfrac{\epsilon}{b(s, k)}
$,
and this can be done
by choosing $n$
large enough.
