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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?

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    $\begingroup$ When did you show that the limit exists? You only assumed the limit exists. $\endgroup$ – Jacky Chong Mar 27 '17 at 5:10
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    $\begingroup$ Hint: $a_2 = 2 \gt1 = a_1$ and then $a_{n+1} \gt a_{n}$ by induction. The sequence is indeed monotonic, but that alone doesn't make it convergent. $\endgroup$ – dxiv Mar 27 '17 at 5:11
  • $\begingroup$ Note that if $a_1 = \frac{1}{2}$, this will be convergent with the limit you calculated. If $a_1$ is above or below this, the limit will diverge to infinity/negative infinity, which you can show by finding the relationship between $a_n$ and $a_{n+1}$ as @dxiv suggests. $\endgroup$ – Mark Mar 27 '17 at 5:15
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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.

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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 $a_n \to \infty$.

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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.

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