Prove the following problem about fibonacci convergence problem

Question

Determine whether the series is convergent or divergent. Give reasons for your answers.

$$\sum_{n=1}^\infty \frac{1}{a_n}$$

where $a_1=a_2=1$ and $a_{n+2}=a_{n+1}+a_n$ for all $n\ge1$.

umm... How do I solve this? I don't even know how to start...

Answer 1

HINT :

An idea is to use the D'Alembert criteria which states that if $$ \frac{a_{n+1}}{a_n} \underset{n \rightarrow +\infty}{\rightarrow} \ell $$ then

• $\ell>1$ makes the series $\sum_{n \in \mathbb{N}}^{ }a_n$ diverge.

• $\ell<1$ makes the series $\sum_{n \in \mathbb{N}}^{ }a_n$ converge.

• $\ell=1$ we cannot conclude whether it converges or not.

Just a simple result before justifying it : if $$ \frac{a_{n+1}}{a_n}\leq k $$ with $k \in \left]0,1\right[$ then the series $\sum_{n \in \mathbb{N}}^{ }a_n$ converges. And if $$ \frac{a_{n+1}}{a_n}\geq k $$ with $k>1$ then the series $\sum_{n \in \mathbb{N}}^{ }a_n$ diverges.

It comes from the fact that we have with the hypothesis $$ a_{n+1} \leq k^n a_0 $$ and the series $\sum k^n$ converges because $k \in \left]0,1\right[$. Same for ther other case $$ a_{n+1} \geq k^n a_0 $$ and the series $\sum k^n$ diverges because $k>1$

Now, let's proof the result I've stated :

By definition of convergence, $\forall \epsilon>0, \ \exists N$ so that for $n>N$ $$ \left|\frac{a_{n+1}}{a_n}-\ell\right|<\epsilon \Rightarrow \ell-\epsilon<\frac{a_{n+1}}{a_n}<\ell+\epsilon $$

• If $\ell>1$, we choose $\displaystyle \epsilon=\frac{\ell-1}{2} \Rightarrow \ell-\epsilon>1$ so with the previous result, the series $\sum_{n \in \mathbb{N}}^{ }a_n$ diverges.

• If $\ell<1$, we choose $\displaystyle \epsilon=\frac{1-\ell}{2} \Rightarrow \ell+\epsilon<1$ so with the previous result, the series $\sum_{n \in \mathbb{N}}^{ }a_n$ converges.

Now, for your exercise, we have $$ a_n=\frac{1}{\sqrt{5}}\left(\phi^n-\left(\overline{\phi}\right)^n\right) $$ where $\displaystyle \phi=\frac{1+\sqrt{5}}{2}>1$ and $\displaystyle \overline{\phi}=\frac{1-\sqrt{5}}{2}<1$ hence $$ \frac{a_{n}}{a_{n+1}}=\frac{\phi^{n}-\left(\overline{\phi}\right)^{n}}{\phi^{n+1}-\left(\overline{\phi}\right)^{n+1}}\underset{(+\infty)}{\sim}\frac{1}{\phi}=\frac{2}{1+\sqrt{5}}<1 $$ Hence with the criteria the series $\displaystyle \sum_{n \in \mathbb{N}^{*}}\frac{1}{a_n}$ converges.

Answer 2

Use the recurrence relation to prove by mathematical induction that $a_{n+1} \ge \dfrac 3 2 a_n$ for $n\ge 3.$ Deduce from that, that $a_n \ge \left(\frac 3 2\right)^{n-3} \cdot 2$ for $n\ge 3.$ Hence $$ \frac 1 {a_n} \le 2 \cdot \left( \frac 2 3 \right)^{n-3} $$ so you have a comparison with a geometric series.