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It is known that : If $\lim_{x \to \infty} \sum_{i=1}^n f_i(x)=0$ and $f_i$ are real-valued periodic function, then $\sum_{i=1}^n f_i(x)=0$ for all $x \in \mathbb{R}$. We can solve this problem by repeatedly using the difference operator $ \Delta_a:(x \rightarrow f(x))\rightarrow (x\rightarrow f(x+a)-f(x))$ and the linearity of this operator.

I try to extend this result to sum of series of periodic function. If $ \lim_{x \to \infty}\sum_{i=1}^{\infty} f_i(x)=0$ and $f_i$ are periodic functions, I want to know under what conditions we can get $\sum_{i=1}^{\infty} f_i(x)=0$.

It is shown in this post that if $f_i$ are merely continuous then the result is not true. On the other hand, I found that if $\sum_{i=k+1}^{\infty}\lVert f_i\rVert =o(2^{-k})$, then the result is true.

Is it true that if the periods of $f_i$ are bounded (i.e. there exist a constant $M$ s.t. $M>\mbox{period of }f_i$ for all $i$) and $\lim_{x \to \infty}\sum_{i=1}^{\infty} f_i(x)=0$, then $\sum_{i=1}^{\infty} f_i(x)=0$?

Besides, if $\sum f_i$ is uniformly convergent and $\lim_{x \rightarrow \infty}\sum_{i=1}^{\infty} f_i(x)=0$, will we have $\sum_{i=1}^{\infty} f_i(x)=0$?

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