MHB Find Limit of Series w/o Quotations: 65 Characters

[IHateFactorial]

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

Find $$\lim_{{n}\to{\infty}}{a}_{n}$$

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:

$$x = \frac{4}{7}x + \frac{3}{7}x = \frac{7}{7}x = x$$

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

$$\lim_{{n}\to{\infty}}{a}_{n} = \infty$$

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:

$$7r^2-4r-3=0$$

$$(7r+3)(r-1)=0$$

And so we know the closed form is:

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

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

$$a_0=c_1+c_2=1$$

$$a_1=-\frac{3}{7}c_1+c_2=2$$

Solving this system, we find:

$$c_a=-\frac{7}{10},\,c_2=\frac{17}{10}$$

Hence:

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

And so we find:

$$\lim_{n\to\infty}a_n=\frac{17}{10}$$