Fibonacci Sequence

The limit we're looking for is simply the golden ratio, φ = 1/2 + √5/2. Thus, the answer is 5.

Of course, this is a very simple exercise if you find the expression of $T_n$ in terms of $n,$ which is $T_n=\frac{(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n}{\sqrt{5}}.$ I am going sketch the solution without using the previous formula.

Let us introduce the notation $c_n=\frac{T_{n+1}}{T_n}.$
Part 1. You can prove by using Mathematical Induction that $c_{n}-c_{n-1}=\frac{(-1)^n}{T_nT_{n-1}}$

Part 2. You can prove that the series $\sum_{n=2}^\infty \frac{(-1)^n}{T_nT_{n-1}}$ converges because it is an alternating series.

Part 3. Since the $(n-1)-$ th partial sum of the series is $c_n-c_2,$ then the sequence ${c_n}$ converges.

Part 4. It is also easy to prove that $c_n=1+\frac{1}{c_{n-1}}$

Part 5 Taking limit on both sides of the previous equation and denoting the $\lim_{n\rightarrow \infty}c_n=c,$ we obtain that $c$ satisfies the equation $c=1+1/c.$

Part 6. Solving the previous equation, we get two values for $c$ : $\frac{1+\sqrt{5}}{2}$ and $\frac{1-\sqrt{5}}{2}.$

Part 7. Since $c$ has to be positive then $c=\frac{1+\sqrt{5}}{2}.$

Then the answer to this problem is $\boxed{5}.$