Determine whether the sequence $a_{n+1} = 3a_n - 1$ is convergent or divergent The sequence is 
$$a_{n+1} = 3a_n - 1$$ where 
$$a_1 = 1$$
Here is my answer:
$$\lim_{n\to\infty}a_{n}=L$$
$$\lim_{n\to\infty}a_{n}=\lim_{n\to\infty}3a_n - 1$$
$$L = 3L - 1$$
$$L = \frac{1}{2}$$
So $a_n$ is convergent and the limit is $\frac{1}{2}$.
Is this sufficient? Do I need to prove that $a_n$ is monotonic?
 A: You've shown that, if the limit exists, then it is equal to $\frac{1}{2}$. But look at the sequence:
$1, 2, 5, 14, 41, 122, 365, \ldots$
Why would that sequence ever equal a half?
A quick application of the ratio test would show that $a_{n+1} > 2a_n$ for any $a_n \geq 1$, so the sequence is bounded from below by $1, 2, 4, 8, 16, \ldots$ which is clearly divergent.
A: You assumed the limit existed (and is finite) and concluded that if it existed it must be $\frac{1}{2}$. You haven't actually showed that existed. But sense you have already put some work in you can continue as follows,
$$a_{n+1}=3a_{n}-1$$
$$\frac{1}{2}=3\left(\frac{1}{2}\right)-1$$
Subtracting the second equation from the first gives,
$$(a_{n+1}-\frac{1}{2})=3(a_{n}-\frac{1}{2})$$
Let $b_n=a_{n}-\frac{1}{2}$ then we have,
$$b_{n+1}=3b_{n}$$
The solution to this is obviously,
$$b_n=b_{1}3^{n-1}$$
So that,
$$a_n=\frac{1}{2}+b_n$$
$$a_n=\frac{1}{2}+(a_1-\frac{1}{2})3^{n-1}$$
Remember we have $a_1=1$.
$$a_n=\frac{1}{2}+\frac{1}{2}(3^{n-1})$$
Clearly as $n \to \infty$ then $a_n \to \infty$.
A: I think you should prove whether the limit exists, before letting $\lim_{n\to\infty} a_n = L$.
This sequence, at a first glance, is divergent. Consider a general case: $a_n, b_n \in {\mathbb R}^m$ ($n = 1,2,\ldots$), $A \in {\mathbb R}^{m\times m}$, and
\begin{equation}
a_{n+1} = A a_n + b_n.
\end{equation}
Then, we have
\begin{align}
a_2 &= A a_1 + b_1\\
a_3 &= A a_2 + b_2 = A^2 a_1 + A b_1 + b_2\\
a_4 &= A a_3 + b_3 = A^3 a_1 + A^2 b_1 + A b_2 + b_3\\
&\ldots\\
a_n &= A^n a_1 +\sum_{k = 1}^{n} A^{n-k}b_k.
\end{align}
For your case, $a_n \in {\mathbb R}$ ($a_1 = 1$), $b_n \equiv -1$, i.e.,
\begin{equation}
a_n = 3^n - \sum_{k = 1}^{n} 3^{n-k} = 3^n\left(1 - \sum_{k = 1}^{n} 3^{-k}\right) \stackrel{(a)}{>} 3^n \frac{1}{2},
\end{equation}
where $(a)$ follows from
\begin{equation}
\sum_{k = 1}^{n} 3^{-k} < \sum_{k = 1}^{\infty} 3^{-k} = \frac{1}{2}.
\end{equation}
Therefore, $a_n$ is unbounded.
PS: It can be also proved that "$a_n$ is convergent iff $a_1 = 1/2$". In that case, $a_n \equiv 1/2$ for all $n = 1,2,\ldots$. Only in that case, your proof $L = 1/2$ works, since the limit exists.
