Limit of sum of periodic function Let $f_1,f_2,...,f_n$ are periodic functions,if $\lim\limits_{x\rightarrow\infty}\sum_{i=1}^n f_i(x)$ is  existent  and bounded. 
How to show $\sum_{i=1}^n f_i(x)\equiv C$ ? 
$C$ is a constant.

I don't think it's true in the full generality of the question.  The questioner needs to give more information about the functions domain an range.  Here's a counterexample.
Define $\chi_n(m) = \begin{cases} 1 , \text{ if } m \in (n), \\ 0, \text{ if } m \notin (n) \end{cases},$
where $(n) \subset \Bbb{Z}$ is the ideal generated by $n \in \Bbb{N}$.  Then each $\chi_n$ is periodic of period $n$.  $\chi_m \chi_n = \chi_{\text{lcm}(m,n)}$ and the characteristic function for $(n) \cup (k)$ is $\phi = 1 - (1-\chi_n)(1- \chi_k) = \chi_n + \chi_k - \chi_k \chi_n$.  In general for the finite set $n= \{(n_i)\}_{i=1..N}$ of ideals we can use the inclusion-exclusion principle to compute $\phi_{\cup n}$.  For instance, $n = \{(a), (b), (c)\}$ and $\phi_{\cup n} = \chi_a + \chi_b + \chi_c - \chi_a\chi_b - \dots + \chi_a \chi_b \chi_c$.
Since $\phi_{\cup n}$ is a finite sum of periodic functions with integer period, its period is simply the lcm of everything which is simply $\text{lcm}(n_1, \dots, n_N)$.
Now let $\psi_n = \phi_{\cup \{(2), \dots, (n)\}}$.  Then $\psi_n$ is a finite partial sum of periodic functions, and
$$
\lim\limits_{n\to \infty} \psi_n (m) = 1
$$
always  for $x \in \Bbb{N}\setminus \{1\}$ (why?), yet its partial sums are not so constant!
