Consider the sequence of functions $f_n(x) =\sum_{k=0}^n a^k \cos(7^k\pi x); |a| < 1$ Show that $f(x) = \lim_{n\to \infty}f _n(x)$ is continuous Consider the sequence of functions
$$f_n(x) =\sum_{k=0}^n a^k \cos(7^k\pi x); |a| < 1$$
Show that
$$f(x) = \lim_{n\to \infty}f _n(x)$$
is a continuous function.
I know that if ${f_n}$ converges uniformly to $f$, then $f$ is continuous.
I also know that ${f_n}$ converges uniformly if $\lim_{n \to \infty}\|f_n - f\|_u = 0$. Hence, I want to show that $\lim_{n \to \infty}\|f_n - f\|_u =\lim_{n \to \infty} \text{sup}\lbrace|f_n - lim_{n\to \infty}f _n(x)|\rbrace = 0 $
I know that the $\cos$ function is between $1$ and $-1$ and hence $f_n$ is between$\sum_{k=0}^n a^k$ and $-\sum_{k=0}^n a^k$, and since both of these sums converge by power series test, $f_n$ must also converge by sandwich theorem. However, I am not sure how to show uniform convergence to $f$ in this instance, which I require for continuity of $f$.
Any help would be appreciated.
 A: You can alternatively use Cauchy's criteria for checking uniform convergence:
A sequence of functions $\{f_n\}$ on $E$ converges uniformly to $f(x)$ in $E$ if and only if given any $\epsilon >0$ there exists an $N$ such that for all $m, n>N$ it is true that $|f_n(x) - f_m(x)| < \epsilon$ for all $x\in E$.
We also have Weierstrass - M test: If $|g_n(x)| \leq M_n$ for all $x\in E$ and for all $n\geq N$ (some $N$), then $\sum g_n(x)$ converges uniformly to some function $g(x)$ if $\sum M_n$ converges.
By Weierstrass M test, the series $\sum\limits_{k=0}^{\infty}a^k\cos(7^k\pi x)$ converges uniformly, since $|a^k \cos(7^k\pi x)| \leq |a|^k$ and $\sum |a|^k$ converges as $|a|<1$. Suppose we denote the partial sum of $\sum\limits_{k=0}^{\infty}a^k\cos(7^k\pi x)$ by $P_n(x)$, then by Cauchy's criteria, given any $\epsilon >0$, there exists an $N$ such that for all $m, n >N$,
$$|P_n(x) - P_m(x)| < \epsilon,$$ for all $x$. Or, for all $x$
$$\Big|\sum\limits_{k=m}^{n} a^k\cos(7^k\pi x)\Big| < \epsilon, $$ where we have assumed that $m\leq n$ without loss of generality. As the $\epsilon$ chosen was arbitrary, what we discussed was that, given any $\epsilon >0$, we can find a $N$ such that for all $m\geq n >N$, and for all $x$,
$$ \Big|\sum\limits_{k=m}^{n} a^k\cos(7^k\pi x)\Big| = |f_n(x) - f_m(x)| <\epsilon.$$
By Cauchy's criteria, it must now imply that $\{f_n(x)\}$ must converge uniformly.
