<s>[INST] Recursive Sequence $a_n = \frac{1}{2} (a_{n-1} + 5) $ I got this question in which they ask me to explain why it is convergent and evaluate its limit. 
$$a_1=3\;and\;a_n = \frac{1}{2} (a_{n-1} + 5) \\ n=2,3,4,... $$
To prove it's convergent, I show that it is increasing and bounded above by 5. Also, I find its limit by showing that 
Let $L=\lim_{n\to\infty} a_n$
Notice that $a_n = \frac{1}{2} (a_{n-1} + 5)$
Hence $\lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{1}{2} (a_{n-1} + 5)$
$\Rightarrow L = \frac{1}{2}(L+5)$ 
$\Rightarrow 2L = L+5$
$\Rightarrow L = 5$
As the sequence is non-decreasing,  $$L =\lim_{n\to\infty} a_n=5 $$

That's what I got. However, the book's answer for this question's limit is $\frac{5}{2}$
Is there anything wrong with my proof?
Thanks in advance.
 [/INST] It might help to solve the recurrence relation
Define $b_n\equiv a_n+c$ and find the value of $c$ that makes $b_n$ a geometric sequence
$$a_n = \frac{1}{2} (a_{n-1} + 5)\implies  b_n-c = \frac{1}{2} (b_{n-1}-c + 5)$$
$$\implies  b_n = \frac{1}{2} b_{n-1}-\frac c2 + \frac 52 +c$$
$$\implies  b_n = \frac{1}{2} b_{n-1} + \frac {c+5}2$$
SO if $c=-5$ we have $ b_n = \frac{1}{2} b_{n-1} $ which is a geometric sequence having solution 
$$ b_n = b_1 \left(  \frac 12\right) ^{n-1}$$
Where $b_1=a_1-5 = 3-5 = -2$
So $$  a_n = b_n + 5 = 5 -2 \left(  \frac 12\right) ^{n-1}=5 - \left(  \frac 12\right) ^{n-2} $$
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