Solving Recurrent Relation

Question

$a_{1}=\dfrac{3}{5}$ , $~$ $a_{n+1}=\sqrt{\dfrac{2a_{n}}{1+a_{n}}}$ $~$ $(n\geq 1)$

Find the closed form of $a_{n}$

Answer 1

Let $b_n = 1/a_n$. Then

$$b_{n+1}^2 = \frac{1+b_n}{2}$$

Sorry if this is kind of a deux ex machina, but you might be able to recognize that the above recurrence fits the pattern of the cosine half-angle formula. That said, the initial condition won't allow for a cosine solution...but it will allow for a hyperbolic cosine solution, which satisfies the same half-angle identity. Then you can write that

$$b_n = \cosh{\left ( \frac{\theta}{2^n}\right)}$$

and

$$a_n = \text{sech}{\left ( \frac{\theta}{2^n}\right)}$$

$$a_1 = \frac{3}{5} \implies \theta = 2 \,\text{arcsech}{\left ( \frac{3}{5}\right)} = 2 \log{3}$$

Therefore

$$a_n = \text{sech}{\left ( \frac{\log{3}}{2^{n-1}}\right)} = \frac{2}{\displaystyle 3^{1/2^{n-1}} + 3^{-1/2^{n-1}}}$$