While reading some things about analytic functions earlier tonight it came to my attention that Fourier series are not necessarily analytic. I used to think one could prove that they *are* analytic using induction

- Let $P(n)$ be some statement parametrized by the natural number $n$ (in this case: the $n$th partial sum of the Fourier series is analytic)
- Show that $P(0)$ is true
- Show that $P(n-1)\Rightarrow P(n)$
- (Invalid) conclusion: $P(n)$ continues to be true as we take the limit $n\to\infty$
^{*}

Why exactly is the conclusion not valid here? It seems very strange that even though $P(n)$ is true for any finite $n$, it ceases to be valid when I remove the explicit upper bound on $n$. Are there circumstances under which I *can* make an argument of this form?

**Example of invalid proof:** Define the truncated Fourier series $F_n(x)$ as the partial sum

$$F_n(x) = \sum_{k=0}^{n} A_k\sin\biggl(\frac{kx}{T}\biggr) + B_k\cos\biggl(\frac{kx}{T}\biggr)$$

where $A_k$ and $B_k$ are the Fourier coefficients for some arbitrary function $f$. Using the facts that $\sin(t)$ and $\cos(t)$ are analytic, and that any linear combination of analytic functions is analytic:

- $P(n)$ is the statement "$F_n(x)$ is analytic"
- $F_0(x)$ is clearly analytic because it is a linear combination of sine and cosine functions
$F_n(x)$ can be written as the linear combination

$$F_{n}(x) = F_{n-1}(x) + A_n\sin\biggl(\frac{nx}{T}\biggr) + B_n\cos\biggl(\frac{nx}{T}\biggr)$$

So if $F_{n-1}(x)$ is analytic, $F_n(x)$ is analytic.

- $F(x) \equiv \lim_{n\to\infty} F_n(x)$ is analytic. But $F(x)$ is the Fourier series for $f$; therefore, the Fourier series for $f$ is analytic.

^{*}I'm assuming that $P(n)$ is a statement about some sequence which is parametrized by $n$ and for which taking the limit as $n\to\infty$ is meaningful