sequence and series (complex analysis) Let $N_0\in \mathbb{N}.$ If a sequence of complex numbers $\{F_N\}_{N \in \mathbb{N}}$ has the following properties:
$$\lim_{N \rightarrow \infty} |F_N|^{1/N}=0$$ and for all $N \geq N_0$,
$$|F_N|\leq \sum_{k=N+1}^{\infty} |F_k|,\quad \quad \quad  $$
then there exists $N_1\in \mathbb{N}$ such that $F_N=0$ for all $N \geq N_1.$
I found this argument on a paper but I cannot prove it. Could you please help me or give me an idea? Thank you.
Masik
 A: For convenience let's say $F_n \geq 0$ for all $n$.  You can compute that if $a < \frac{1}{2}$ then $a^N > \sum_{n=N+1}^\infty a^n$.  Since $F_n$ is going to $0$ more quickly than $2^{-n}$ we would expect that  $$F_N > \sum_{n=N+1}^\infty F_n$$ for large $N$ unless perhaps $F_n$ is eventually $0$.  Let's try to make this rigorous.
Since $F_n^{1/n} \rightarrow 0$, for each $\epsilon >0$ there is an $M$ so that $F_n < \epsilon^n$ for $n > M$.  For each $\epsilon$, let $M_\epsilon$ be the smallest such $M$.  Thus if $M_\epsilon > 1$, so that there is at least one $n$ with $F_n \geq \epsilon^n$, we have $F_{M_\epsilon} \geq \epsilon^{M_\epsilon}$ but $F_n < \epsilon^n$ for $n > M_\epsilon$.
Now suppose that $F_n$ is not eventually $0$.  Then for every $N$ there is some $n>N$ with $F_n >0$; in particular, setting $\epsilon = F_n$ we see that $M_\epsilon > N$. This tells us that $M_\epsilon \rightarrow \infty$ as $\epsilon \rightarrow 0$.
Now choose $0 < \epsilon < \frac{1}{2}$ so that $M_\epsilon > N_0$.  Then we have 
\begin{align*}
\epsilon^{M_\epsilon} \leq F_{M_\epsilon} \leq\sum_{n= M_\epsilon +1}^\infty F_n < \sum_{n= M_\epsilon +1}^\infty \epsilon^n = \epsilon^{M_\epsilon}\frac{\epsilon}{1-\epsilon} < \epsilon^{M_\epsilon}
\end{align*}
a contradiction.
A: This may be another way to do it. 
So given all the conditions, there exists $M$ such that $\forall N\ge M, |F_N|^{1/N}<1/3,$ and $|F_N|\le\sum_{k=N+1}^\infty|F_k|,$ I claim that for those $N$'s, $|F_N|\le(1/2)^n(1/3)^N$ for any non-negative integer $n$. Then by letting $n$ approaches $\infty,$ we actually force $|F_N|=0.$
To prove the claim, we do induction on $n.$ When $n=0,$ it follows immediately from $|F_N|^{1/N}<1/3.$ In general, by induction hypothesis 
$$|F_N|\le\sum_{k=N+1}^\infty|F_k|\le\sum_{k=N+1}^{\infty}(1/2)^n(1/3)^k=(1/2)^{n+1}(1/3)^N$$
