Is there any sequence $\{a_n:n\in \mathbb N\}$ with each non-zero terms s.t. $\displaystyle \sum_{n=1}^{\infty}a_n=0$ 
Is there any sequence $\{a_n:n\in \mathbb N\}$ with each non-zero terms s.t. $\displaystyle \sum_{n=1}^{\infty}a_n=0$?

I'm finding an infinite linearly dependent set which has no finite linearly dependet subset. If I could find the above then I could say so.
Edit: sorry I wrote first 'strictly positive' terms instead of 'non-zero'.
 A: No.If there were one as you said then $0 < a_1 \le \sum a_n = 0$, contradiction. 
A: No, there is not such sequence. If there were a sequence of nonnegative terms $a_n$ with one strictly positive term $a_{n_0}$ then we would have
$$
\sum_{n=1}^\infty a_n\geq a_{n_0}>0.
$$

[For the updated question] Yes: If $\{a_n:n\geq1\}$ is any sequence of non-zero terms such that $S:=\sum_{n=1}^\infty a_n\neq0$ then the sequence $\{b_n:n\geq1\}$ defined by $b_1=S,\ b_n=a_{n-1}$ is a sequence of non-zero terms and its sum is $0$.
A: Let convergence be understood in the $\|\cdot\|_\infty$ norm.
Consider the familiy of sequences $$\mathcal{F} = \left\{-\left(\underbrace{0, 0, \ldots, 0}_{n-1}, \frac{1}{2^n}, 0, 0, \ldots\right) : n \in \mathbb{N}\right\} \cup \left\{\left(1, \frac{1}{2}, \frac{1}{4}, \ldots, \frac{1}{2^n}, \frac{1}{2^{n+1}}, \ldots\right)\right\} \subseteq c_0$$
Notice that every finite subset of $\mathcal{F}$ is linearly independent, since the last sequence has infinite support, and the others have finite disjoint supports.
Also,
$$\sum_{x \in \mathcal{F}}x = \left(1, \frac{1}{2}, \frac{1}{4}, \ldots, \frac{1}{2^n}, \frac{1}{2^{n+1}}, \ldots\right) - \sum_{n=1}^\infty\left(\underbrace{0, 0, \ldots, 0}_{n-1}, \frac{1}{2^n}, 0, 0, \ldots\right) = 0$$
since:
$$\left\|\left(1, \frac{1}{2}, \frac{1}{4}, \ldots, \frac{1}{2^n}, \frac{1}{2^{n+1}}, \ldots\right) - \sum_{n=1}^N\left(\underbrace{0, 0, \ldots, 0}_{n-1}, \frac{1}{2^n}, 0, 0, \ldots\right)\right\|_\infty$$
$$=\left\|\left(1, \frac{1}{2}, \frac{1}{4}, \ldots, \frac{1}{2^n}, \frac{1}{2^{n+1}}, \ldots\right) - \left(1, \frac{1}{2}, \frac{1}{4}, \ldots, \frac{1}{2^{N}}, 0, 0, \ldots\right)\right\|_\infty$$
$$=\left\|\left(0, 0, \ldots, 0, \frac{1}{2^{N+1}}, \frac{1}{2^{N+2}}, \ldots\right)\right\|_\infty = \frac{1}{2^{N+1}}\xrightarrow{N\to\infty} 0$$
This property is sometimes called $\omega$-independence:
$$\sum_{n=1}^\infty\alpha_n a_n = 0 \implies \alpha_n = 0, \,\forall n\in\mathbb{N}$$
Thus, $\mathcal{F}$ is (finitely) linearly independent, but it is not $\omega$-independent.
