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?

  • 2
    $\begingroup$ When did you show that the limit exists? You only assumed the limit exists. $\endgroup$ – Jacky Chong Mar 27 '17 at 5:10
  • 1
    $\begingroup$ 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

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.

| cite | improve this answer | |

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,



Subtracting the second equation from the first gives,


Let $b_n=a_{n}-\frac{1}{2}$ then we have,


The solution to this is obviously,


So that,



Remember we have $a_1=1$.


Clearly as $n \to \infty$ then $a_n \to \infty$.

| cite | improve this answer | |

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.