Find Limit of Series w/o Quotations: 65 Characters

[IHateFactorial]

Let \(\displaystyle {a}_{n+1} = \frac{4}{7}{a}_{n} + \frac{3}{7}{a}_{n-1}\) where a₀ = 1, and a₁ = 2.

Find \(\displaystyle \lim_{{n}\to{\infty}}{a}_{n}\)

Well, seeing as it says that x approaches infinity, the difference between where points a_n-1, a_n, and a_n+1 are plotted on the y-axis is almost insignificant, so we can simply apply a common value of x to all a_is in the function. It would become:

\(\displaystyle x = \frac{4}{7}x + \frac{3}{7}x = \frac{7}{7}x = x\)

Seeing as x equals itself, the higher the value of n, we can say that:

\(\displaystyle \lim_{{n}\to{\infty}}{a}_{n} = \infty\)

Is this right? Or did I screw it up?

 

[MarkFL]

What I would do here is look at the characteristic equation for the linear homogeneous recursion:

\(\displaystyle 7r^2-4r-3=0\)

\(\displaystyle (7r+3)(r-1)=0\)

And so we know the closed form is:

\(\displaystyle a_n=c_1\left(-\frac{3}{7}\right)^n+c_2\)

And we can use the initial values to determine the parameters:

\(\displaystyle a_0=c_1+c_2=1\)

\(\displaystyle a_1=-\frac{3}{7}c_1+c_2=2\)

Solving this system, we find:

\(\displaystyle c_a=-\frac{7}{10},\,c_2=\frac{17}{10}\)

Hence:

\(\displaystyle a_n=\frac{17-7\left(-\dfrac{3}{7}\right)^n}{10}\)

And so we find:

\(\displaystyle \lim_{n\to\infty}a_n=\frac{17}{10}\)