Nth term of recursively defined sequence

Question

I have the following sequence defined recursively: $s_{n+1}=1+\frac{s_{n}}{2}$. I understand that the nth term of this sequence is given by $f(n)=2-(2-t)2^{1-n}$. However, I am not sure how this conclusion is reached (in particular, how this general term is derived from the recursive definition). Could someone provide a hint (if not demonstrate) how this is done? Thank you.

Answer 1

$s_{n+1}=1+\frac{s_{n}}{2}$ tells that $s_{n+1}$ is essentially $\frac{s_{n}}{2}$.

If that were true exactly, then we'd have $s_n = s_0 2^{-n}$.

Perhaps we can transform $s_n$ to $u_n$ such that $u_{n+1}=\frac{u_{n}}{2}$. Try $u_n=s_n+a$ for some $a$.

Answer 2

One general way uses generating functions. Define $S(z) = \sum_{n \ge 0} s_n z^n$, multiply your recurrence by $z^n$ and sum over $n \ge 0$, recognize the resulting sums:

$\begin{align*} s_{n + 1} &= \frac{1}{2} s_n + 1 \\ \sum_{n \ge 0} s_{n + 1} z^n &= \frac{1}{2} \sum_{n \ge 0} s_n z^n + \sum_{n \ge 0} z^n \\ \frac{S(z) - s_0}{z} &= \frac{1}{2} S(z) + \frac{1}{1 - z} \end{align*}$

You don't give the initial value, just call it $s_0$. Solve for $S(z)$, write as partial fractions (a computer algebra system is nice to have...):

$\begin{align*} S(z) &= \frac{2 s_0 - 2 (s_0 - 1) z}{2 - 3 z + z^2} \\ &= \frac{2}{1 - z} + \frac{s_0 - 2}{1 - z / 2} \end{align*}$

We want the coefficient of $z^n$ in this. Just two geometric series:

$\begin{align*} [z^n] S(z) &= 2 + (s_0 - 2) \cdot 2^{-n} \end{align*}$