Linearity of fourier series

Question

I am given the following function:

$$ g(x) = \begin{cases} sin(\frac{\pi^2}{x}), & x \in[-\pi,\pi),x\neq0 \\ 0, & x=0 \end{cases}$$

It's fourier series is: $\sum_{n=1}^\infty b_nsin(nx)$

Now I am trying to find the fourier series of $f(x) :=g(x)+1$.

I want to say that the series is $1+\sum_{n=1}^\infty b_nsin(nx)$, but why is this allowed? I know that $\hat f(n)+\hat g(n) = \widehat {(f+g)}(n)$ ($\hat f(n)$ denotes the n fourier coeffiecient of f). But can I conclude that I am allowed to add the series ? or is it only allowed in case both of the series absolutely converge?

Edit: Basically that I'm trying to prove is the following:

$$\sum_{n \in \Bbb Z} (\hat g(n)+1)e^{inx} = \sum_{n \in \Bbb Z} \hat g(n)e^{inx}+\sum_{n \in \Bbb Z}1e^{inx} = \sum_{n \in \Bbb Z}\hat g(n)e^{inx}+1$$

(The first quailty is what I don't get)

Answer 1

1 is orthogonal to all the terms $\sin(nx)$. In fact, it is one of the basis elements for Fourier series $\{1,\cos(nx),\sin(nx) \}$, which form an orthogonal basis.

Edit: Your equations are incorrect. It should read as follows, with f(x)=1

\begin{equation} \sum\widehat{(g+f)}e^{inx} = \sum(\widehat{g}+\widehat{f})e^{inx} = \sum\widehat{g}e^{inx} +\sum\widehat{f}e^{inx} = \sum b_n \sin(nx) + 1 \end{equation}

Note that the only nonzero Fourier coefficient of 1 is the constant term.

Answer 2

The type of Fourier series is no longer the same because you no longer have the same boundary condition. Accordingly you become forced to include a constant term as one of your basis functions. But unlike what you wrote, making this change does not add $1$ to each Fourier coefficient. It just adds in a term, say $\phi_0(x)=1$ so that $b_0=1$.