I would like to have your help about the proof of the following statement:
If the sequences of complex numbers $\{F_N\}_{N \in \mathbb{N}},\{G_N\}_{N \in \mathbb{N}}$ have the following properties:
(i) $\lim_{N \rightarrow \infty} |F_N|^{1/N}=0;$
(ii) $\limsup_{N \rightarrow \infty} |G_N|^{1/N}\leq 1;$
(iii) there exists $N_0\in \mathbb{N}$ and $C>0$ such that $|F_N|\leq C|G_N| \sum_{k=N+1}^{\infty} |F_k|,$ for all $N \geq N_0$,
then there exists $N_1\in \mathbb{N}$ such that $F_N=0$ for all $N \geq N_1.$
This statement is in a paper without proof. Moreover, I only know the prove of this statement for the case when $G_N=1$ for all $N.$
The following is the proof for this special case.
Given the assumptions, there exists $M$ such that for all $ N\ge M,$ $$ |F_N|^{1/N}<\frac{1}{C+2}, \quad \quad \quad \quad |F_N|\le C\sum_{k=N+1}^\infty|F_k|.$$ We claim that for those $N$'s, $$|F_N|\le\left(\frac{C}{C+1}\right)^n\left(\frac{1}{C+2}\right)^N$$ for any non-negative integer $n$. Then, letting $n\rightarrow \infty,$ we see that $|F_N|=0.$
To prove the claim, we use induction on $n.$ When $n=0,$ the formula follows immediately from $|F_N|^{1/N}<1/(C+2).$ In general, using induction it follows that $$|F_N|\le C\sum_{k=N+1}^\infty|F_k|\le C\sum_{k=N+1}^{\infty}\left(\frac{C}{C+1}\right)^n\left(\frac{1}{C+2}\right)^k=\left(\frac{C}{C+1}\right)^{n+1}\left(\frac{1}{C+2}\right)^N.$$
Could you provide me a proof of the statement or give me an idea of the proof? Thank you very much.
Masik