Guess a formula for $a_n$ and proof it by induction.

Question

Series $a_1, a_2,a_3…$ is defined for all $n≥1 ,a_1= 1$ and after that $a_{n+1} = a_n/(2+a_n )$, for n≥1. Guess a formula for $a_n$ and proof it by induction. I would need help with this

Answer 1

I calculated the first $6$ terms: $$a_1 = 1,\ a_2 = \frac{1}{3},\ a_3 = \frac{1}{7},\ a_4 = \frac{1}{15},\ a_5 = \frac{1}{31},\ a_6 = \frac{1}{63}$$ From this, we can have a pretty good guess that $$a_n = \frac{1}{2^n-1}$$ Proof by induction:

I. It works for $n=1$: $$a_1 = \frac{1}{2^1-1} = \frac{1}{1} = 1$$ II. Assume that $$\forall k \in \{1,\dots,n\}:\ \ a_k = \frac{1}{2^k-1}$$ III. Prove for $k+1:$ $$a_{k+1} = \frac{a_k}{2+a_k} = \frac{\frac{1}{2^k-1}}{2+\frac{1}{2^k-1}}=\frac{1}{2^k-1}\cdot\frac{1}{\frac{2^{k+1}-2+1}{2^{k}-1}}=\frac{1}{2^k-1}\cdot\frac{2^{k}-1}{2^{k+1}-1} = \frac{1}{2^{k+1}-1}.$$

Answer 2

Hint: Consider that $$ \frac{a_n}{2+a_n}=\frac{1}{\dfrac{2}{a_n}+1} $$ so $$ \frac{1}{a_{n+1}}=\frac{2}{a_n}+1 $$

Can you make a guess for $b_n$ defined by $b_1=1$ and $b_{n+1}=2b_n+1$?