Proving that this sequence is convergent Given $a_1 = 2$, and $ a_{n+1} = \frac{a_n+5}{4} $ for all $n > 1$ , is this sequence convergent? Give a formal proof in either case (converges or diverges).
Attempt: I do think this converges, but cannot say for sure. 
$a_1 = 2$
$a_2 = \frac{7}{4}$
$a_3 = \frac{27}{16}$
$a_4 = \frac{107}{64}$
and so on. I can see that it is decreasing and seems to be bounded below by something. But I do not know how to present it formally. 
Any help?
 A: The sequence is descendant and positive, so it's convergent.
Where does it converge to ? Supposedly $l$ is its limit.
$l=\frac{l+5}{4}$ gives $l=\frac{5}{3}$
A: Consider
$$
\begin{align}
a_{n+1}-\frac53
&=\frac{a_n+5}4-\frac53\\
&=\frac{a_n-\frac53}4
\end{align}
$$
Can you find a formula for $a_n-\frac53$?
A: Try to show that the sequence is bounded and strictly decreasing , this will imply convergence. Hint use induction 
A: Let $x \in \mathbb R$. If  $x \gt 5/3$ then we always have
$\tag 1    \frac{5}{3} \lt \frac{x + 5}{4} \lt x$
Since our sequence $a_n$ is strictly decreasing and always greater than $5/3$, it has a limit which we denote by $\alpha$.
For the sequence to converge, $(a_{n+1} - a_n)$ must converge to zero, and
$\quad a_{n+1} - a_n =  \frac{a_n + 5}{4} - a_n = \frac{-3 a_n + 5}{4} $
and applying the limit operation,
$\quad \lim_{n\to +\infty} \frac{-3 a_n + 5}{4} = \frac{-3 \alpha + 5}{4} = 0$
Solving we find the limit of the sequence is equal to $5/3$.
A: The formal proof that the OP asked for can be found by combining his work with the (partial) answers given by Abra001 and robjohn.
If the sequence converges, then the sequence $(b_n)$ defined by $b_n = a_{n+1}$ converges to the same limit. So if the limit exists, it can only be $5/3$, as we see in the answer provided by Abra001.
As robjohn points out, everything might fall into place if we can get a closed formula. He encourages the OP to find a pattern.
The OP provided us with an interesting pattern, but we might not be able to look at it and come up with a formula. But perhaps we should see if we can get a formula for a 'nicer' sequence, $c_n = a_n - 5/3$. Our 'jump-start' approach tells us that it converges to $0$. Robjohn points the way by writing out some algebra.
He also has a comment that shows a table. But, we can also just crank out the numbers if we want to avoid the algebra. In fact, if you hate working with fractions you can run a computer program.
Consider this:
# Python Program:

from fractions import Fraction

a = Fraction(2,1)
for i in range(5):
    b = a - Fraction(5,3)
    print(b)
    a = Fraction(a + 5, 4)


OUTPUT:
1/3
1/12
1/48
1/192
1/768

We've found the pattern!
$\tag 1 a_n = \frac{5}{3} + \frac{1}{3 \, 4^{(n-1)}} $
Now for the formal proof:
State that it not difficult to show that $\text{(1)}$ holds. Alternatively, verify it using induction and algebra.
State that the convergence of the sequence $(a_n)$ to $5/3$ is now trivial, using $\text{(1)}$.

Note: The above might be 'overkill', since the OP only wanted to show that the sequence converges, which we know once we show that $a_{n+1} \lt a_n$ for all $n \ge 1$.
A: Hint: try computing $a_{n+1}/a_n$ and using a convergence theorem
A: Hint: Use induction to show it's decreasing. Since it's positive it's bounded from below, thus it converges. Say it converges $L \in \mathbb{R}$. Apply the limit on both sides of the recursive formula of the sequence to find the value of L
A: $$a_2=\frac{a_1}{4} + \frac{5}{4}\qquad a_3=\frac{a_1}{4^2}+\frac{5}{4^2}+\frac{5}{4} \qquad a_4=\frac{a_1}{4^3}+\frac{5}{4^3}+\frac{5}{4^2}+\frac{5}{4}$$
Get an explicit formula for $a_n$ by guessing from this pattern, proof it by induction, then proof that the explicit formula converges or diverges.
