TITLE: Convergence of a recurrence
QUESTION [6 upvotes]: Given the recursive definition (starting with a positive integer) 
$$
a_n = \frac{a_{n-1}}{2}+4
$$
I am trying to find an explicit form and show that it approaches 8.  So I started by writing it out, starting with an arbitrary value $a_1$, and one way to write this is
$$
\begin{matrix}
\frac{\frac{\frac{a_1}{2}+4 }{2}+4}{2}+4
&  \\
& \ddots
\end{matrix}
$$
or writing it term by term we could have
$$
\begin{matrix}
  \frac{a_1}{2}+4 \\
  \frac{a_1}{4}+6 \\
  \frac{a_1}{8}+7 \\
  \frac{a_1}{16}+7.5 \\
  \frac{a_1}{32}+7.75 \\
\end{matrix}
$$
From the pattern it seems that an explicit formula for the $k^{th}$ term would be
$$
a_k = \frac{a_1}{2^k} + \left(8 - \frac{4}{2^{k-1}}\right)
$$
which we could rewrite as 
$$
a_k = 8 + \frac{a_1-8}{2^{k}}
$$
and then to show convergence it is just
$$
\lim_{k \rightarrow \infty} \left[ 8 + \frac{a_1-8}{2^{k}} \right]= 8
$$
This seems to work, but I simply looked for a pattern and guessed the explicit formula.  My question is if there is a direct way to get from the recursive definition to the explicit one.

REPLY [2 votes]: Simple trick, if its going to converge on something, then we must have $\lim_{n→\infty}(a_{n+1}=a_n)$ ie the definition of converging