Uniform convergence of the Fourier Series using Bessel's inequality Consider the Fourier series of $f$, 
$$ \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(nx) + b_n \sin(nx) $$
Let $$f_n(x)= a_n \cos(nx) + b_n \sin(nx)$$
Then to show that $f_n(x)$ is uniformly convergent we must show that $\forall \epsilon >0$, there exists $N$ such that $|\sum_{n=k}^{l} f_n(x) | < \epsilon$ whenever $N < k < l$.
Then $$\big|\sum_{n=k}^{l} f_n(x) \big|\le \sum_{n=k}^{l} |f_n(x)| \le \sum_{n=k}^{l} |a_n| + |b_n| $$
From previous work I have found,
$$a_n'=n\cdot b_n$ an $b'_n=-n\cdot a_n$$
Since $\sum_{n=k}^{l} |a_n|$ and $\sum_{n=k}^{l} |b_n|$ are monotone we can show they converge by proving boundedness. 
Then $$\sum_{n=1}^{N} |a_n| = \sum_{n=1}^{N} | \frac{1}{n}| |b_n'|$$
$\star$ We then use Cauchy-schwarz to show that $\sum_{n=1}^{N} | \frac{1}{n}| |b_n'| \le \sqrt{\sum \frac{1}{n^2}}\sqrt{\sum b_n'^{2}}$.
The proof is concluded by the statements that boundedness follows from Bessel's inequality.
$\textbf{Question:}$ I have the Cauchy-Schwarz inequality written as follows $<a,b> \le |a||b|$. I do not see how $\sum_{n=1}^{N} | \frac{1}{n}| |b_n'| \le \sqrt{\sum \frac{1}{n^2}}\sqrt{\sum b_n'^{2}}$ relates to this?
$\textbf{Question:}$ How do I use Bessel's inequality to show that $sqrt{\sum b_n'^{2}}$ is finite?,
 A: For the first question, the Cauchy-Schwarz inequality holds for every inner product, but if we restrict ourselves to the finite sums here, we can do with the standard inner product on $\mathbb{R}^N$. The sum
$$\sum_{n=1}^N \frac{1}{n}\lvert b_n'\rvert$$
is the inner product of the vector $x = (1,1/2,\dotsc, 1/N)$ whose $n$-th coordinate is $1/n$, and the vector $y = (\lvert b_1'\rvert,\lvert b_2'\rvert,\dotsc,\lvert b_N'\rvert)$ whose $n$-th coordinate is $\lvert b_n'\rvert$. The norm of $x$ is
$$\lvert x\rvert = \sqrt{\sum_{n=1}^N \frac{1}{n^2}},$$
and the norm of $y$ is
$$\lvert y\rvert = \sqrt{\sum_{n=1}^N \lvert b_n'\rvert^2} = \sqrt{\sum_{n=1}^N b_n'^2},$$
provided the $b_n'$ are real, what they are when $f$ (and hence $f'$) is real-valued. Thus the Cauchy-Schwarz inequality is here precisely
$$\sum_{n=1}^N \frac{1}{n}\lvert b_n'\rvert = \langle x,y\rangle \leqslant \lvert x\rvert\,\lvert y\rvert = \sqrt{\sum_{n=1}^N \frac{1}{n^2}}\sqrt{\sum_{n=1}^N b_n'^2}.$$
For the second question, Bessel's inequality states
$$\frac{a_0^2}{2} + \sum_{n=1}^N \bigl( a_n^2 + b_n^2\bigr) \leqslant \frac{1}{\pi}\int_0^{2\pi} g(x)^2\,dx$$
for all $N$ and all Riemann integrable real-valued functions $g$ defined on $[0,2\pi]$ (or, more generally, if you have the background, for all square-integrable $g$), where the $a_n$ resp. $b_n$ are the Fourier coefficients of $g$. Since the right hand side is a finite constant, the sums $\sum\limits_{n=1}^N b_n^2$ are bounded (as are the $\sum a_n^2$ sums).
For a piecewise continuously differentiable $f$, the derivative $f'$ is piecewise continuous, hence bounded and Riemann integrable, so the boundedness of $\sum\limits_{n=1}^N b_n'^2$ follows from Bessel's inequality.
