Is there any necessary and sufficient condition for function $f$ such that:


$f(x)=\sum\_{k=1}^{\infty} f\_k(x)$ for all $x \in \mathbb{R}$,where $(f\_n )\_{n=1}^{\infty}$ is a sequence of periodic function on $\mathbb{R}$ ??


(note that $f\_n$ may not be integrable or measurable)


Besides,it is known that $ \lim\_{x \rightarrow \infty} \sum\_{k=1}^{n}$$f\_k(x)=0$ implies 
$\sum\_{k=1}^{n}$$f\_k(x)=0$.


I wonder if it is also true that $\lim\_{x \rightarrow \infty} \sum\_{k=1}^{\infty}$$f\_k(x)=0$ implies 
$\sum\_{k=1}^{\infty}$$f\_k(x)=0$.



represented as a series of periodic function
Every function can be written as the pointwise sum of a sequence of periodic functions. 


Given $f$. Let $f\_1 = f$ on $(-2,2]$ and extend as a function of period 4. And follow the following recursive definition:


Let $ f\_{k+1}(x) = f(x) - \sum\_{n= 1}^k f\_k(x)$ for $x\in (-2^{k+1},2^{k+1}]$ and extend as a function of period $2^{k+2}$. 


Observe that if $|x| < 2^{k}$, by definition $f\_j(x) = 0$ for all $j > k$. Hence the series converges for all $x$ pointwise. 




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Note that the above construction also gives a counterexample to your final question: just start with any $f$ not identically zero with compact support. 


Of course, the construction given does not converge uniformly (quite far from it, usually). And the answer will likely be different if you impose an additional requirement that the periods of the various functions $f\_k$ remain bounded. 


