Question

Determine whether \(a_n\) is converging or diverging
\(a_1\)=2
\(a_{n+1}\)=3\(a_n\)-2

Answer 1

\[
\begin{align*}
a(n+1) &= 3a(n) - 2\\
&=3( 3 a(n-1) - 2) - 2 \\
&= 3^2 (3 a(n-2) - 2) - 2( 1 + 3)\\
&= 3^3 a(n-2) - 2(1 + 3 + 3^2)\\
&= \dots \\
&= 3^na(n-(n-1)) - 2(1 + 3 + \dots + 3^{n-1}) \\
&= 3^n a(1) - 2 \frac{3^{n} - 1}{3 - 1} \\
&= 2 \cdot 3^n - 3^n + 1 \\
&= 3^n + 1 \\
\end{align*}
\]

let's see if I can find any nicer methods.

Answer 2

just for convergence,
\[
\sum_{k=1}^n (a(k+1) - a(k))= a(n+1) - a(1) = \sum_{k=1}^n 2 (a(k) - 1 ) = 2 \sum_{k=1}^n (a(k) - 1)\]
By induction we prove that \( a(k) > 2\) which is enough to show that the sequence diverges.

Answer 3

Let me explain the second method a bit
\[ a(n+1) = 3 a(n) - 2 \implies a(n+1) - a(n) = 2 (a(n) - 1 ) \]
then summing we get
\[ \sum_{k=1}^n (a(k+1) - a(k)) \\ = a(n+1) - a(n) + a(n) - a(n-1) + a(n-1) + \dots + a(2) - a(1) \\ = a(n+1) - a(1) \]
but we also know the the difference of consecutive terms.
\[
a(n+1) - a(1) = \sum_{k=1}^n 2 (a(k) - 1) = 2 \sum_{k=1}^n (a_k - 1 )
\]

We know each \( a_1 = 2, a_2 > 2,a_3>2, ... \) so we use induction to sow that this is true for all \( n> 1\)

let \( a_k > 1 \) then
\[

a(k+1) = 3 a(k) - 2 > 3\cdot 2 - 2 = 4
\]
So all \( a_n >2 \forall n>1 \) which gives inequality
\[
a(n+1) - 2 > 2 \sum_{k=1}^n n
\]

we know the right diverges so left is even \(>\) right so left diverges.

Answer 4

Here is another way,
\[
\begin{align*}
a_{n+1} &= 3 a_n - 2 \hspace{1cm}(a) \\
a_n &= 3a_{n-1} - 2 \hspace{1cm}(b) \\
(a) - (b) &\implies a_{n+1} - a_n = 3a_n - 3a_{n-1 } \\
a_{n+1} &= 4 a_n - 3 a_{n-1}
\end{align*}
\]

this is a homogeneous linear difference equation. There are couple of methods for solving this. One is via using matrix it has been discussed in this video lecture (might be above/below it)

other is via discrete z-transform.
http://en.wikipedia.org/wiki/Z-transform
it assumes that there exits a number \(r \) such that \(r^n = a_n\) , if you can find \(r\) such that it is satisfied, then it's fine (i don't know how) ... just like solving second order constant coefficients (although it can be done differently)
this gives you
\[ r^{n+1} = 4 r^n - 3 r^{n-1}\]
on simplifying you get
http://www.wolframalpha.com/input/?i=r^2+%3D+4r+-+3
so your solution is of the form
\[ a_n = c_1 r_1^n + c_2 r_2^n \]
where \(r_1\) and \(r_2\) are roots of above. and c's are constant to be determined.

putting \(a_1 = 2\) and \(a_2 = 4\) you get
\( c_1 = 1/3 \) and \( c_2 = 1 \)
http://www.wolframalpha.com/input/?i=a%28n%2B1%29+%3D+4+a%28n%29+-+3+a%28n-1%29%2C+a%281%29+%3D+2%2C+a%282%29+%3D+4

Answer 5

the video
http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/lecture-23-differential-equations-and-exp-at/

Answer 6

*just like solving second order differential equations with constant coefficients