There are already proofs for this on Stackexchange, but I am seeking a specific one I am interested in.
Let $\{u_k\}^\infty_{k=0}$ be the Fibonacci sequence, where $u_0=u_1=1$ and $u_{k+1}=u_k+u_{k-1}$. Then $\sum_{k=1}^{\infty} \tfrac{1}{u_k}$ is convergent. I want to prove it with the following two statements:
For $a_k≠0, \forall k \in \mathbb{N}$, if $$\limsup_{k\to\infty}\left|\frac{a_{k+2}}{a_k}\right|<1$$ then the series $\sum_{k=1}^\infty a_k$ is absolutely convergent.
$\limsup_{k\to\infty}\left|\frac{u_{k-1}}{u_{k+1}}\right|<1$
Can anyone help me?
First notice that $u_k \geq 1$ for all $k$ and $\{u_k\}$ is increasing monotonously. From $u_{k+1}=u_k+u_{k-1}$ and $u_k \geq u_{k-1}$ follows, that $$u_{k+1} \geq 2u_{k-1} $$ and thus $$\frac{a_{k+1}}{a_{k-1}}=\frac{u_{k-1}}{u_{k+1}} \leq \frac{1}{2},$$ whereas $a_k=\frac{1}{u_k}$. From this we can deduce that
$$\limsup_{k\to\infty}\left|\frac{a_{k+1}}{a_{k-1}}\right|\leq \frac{1}{2}.$$
From the fact that $\sum a_k$ converges if $\limsup_{k\to\infty}\left|\frac{a_{k+2}}{a_{k}}\right|<1$, we can see that $\sum \frac{1}{u_k}$ converges.