I know that the rate of convergence $\mu$ of a convergent sequence $(x_n)_{n\in\mathbb N} \to x$ in $\mathbb R$ is given by $$\mu = \lim_{n\to\infty}\frac{|x_{n+1} - x|}{|x_n - x|}.$$ But if I have a sequence of functions $(f_n(x))_{n\in\mathbb N} \to f(x)$, is there an analogous way one can determine the rate at which $f_n$ converges to $f$? In particular, I am interested in the Fourier series $f_n, g_n : [0,1] \to \mathbb R$ given by
$$ f_n(x) = - \sum_{k = 1}^n \frac{2(-1)^k}{k\pi} \sin (k\pi x) \qquad \text{and} \qquad g_n(x) = \frac{1}{2} + \sum_{k = 1}^n\frac{2((-1)^k-1)}{k^2 \pi^2} \cos (k\pi x)$$
where both $f_n$ and $g_n$ converge to $f(x) = x$ for $x\in [0,1]$.
Experimentally (through different plots for different values of $n$), I know that $f_n$ converges to $f$ more slowly than $g_n$ does, but I do not know how to prove this rigorously. I appreciate any assistance.
I've decided to try and answer my own question. First of all, as has been mentioned in the comments, $f_n \to f$ only pointwise, whereas $g \to f$ uniformly. This already indicates that $g_n$ converges more strongly.
Now treating $f$ and $g$ as two elements of the function space $\mathbb R ^ {[0,1]}$ with the $L^2$ norm $$\left\| \phi \right\| = \int_0^1 |\phi(x)|^2 \,\mathrm d x,$$ we define the rate of convergence of $\phi_n$ to $\phi$ by $$\mu_{\phi_n,\,\phi} = \lim_{n \to \infty} \frac{\left\| \phi_{n+1}-\phi \right\|}{\left\| \phi_n - \phi \right\|}.$$
What we want to show is that $\mu_{f_n,\,f} < \mu_{g_n, \, f}$. Using Paversal's identity, we have that $$ \|\phi\|^2 = \sum_{n} |\alpha_n|^2,$$ where $\alpha_n$ are the Fourier coefficients of $\phi$. In our case, we have $$\| f - f_n \|^2 = \sum_{k=n+1}^\infty |\alpha_k|^2 = \sum_{k=n+1}^\infty \frac{|a_k|}{k^2}.$$ This gives an expression of the form $ (c - H_n^{(2)})/(d - H_n^{(4)})$ where $H_n^{(2)}$ denotes the generalised harmonic number, and c,d are the corresponding $H_\infty$ terms. So the question boils down to showing that $\sum_k 1/k^2$ converges more slowly than $\sum_k 1/k^4$, which can be done easily using the original definition of rate of convergence.