How to find limit of the sequence given by $f_{n+1}=\frac{3}{7}f_n+8$

Question

A sequence is defined by recurrence relation $f_{n+1}=\frac{3}{7}f_n+8$ with $\mu_0=-14$, then what is the limit of the sequence?

1. $14$
2. $-14$
3. $\frac{-3}{7}$
4. $\frac{3}{7}$

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My attempt:

As wiki : the limit of a sequence is the value that the terms of a sequence "tend to". If such a limit exists, the sequence is called convergent.

I'm stuck here what is $\mu_0=-14$? How do I find limit of the sequence?

Can you explain in formal way, please?

Answer 1

$\mu_0$ is probably $f_0$, the first term of the sequence. As already said, if the limit exists ($\exists \lim_{n \to \infty}f_n = F$), then we can write $$\lim_{n \to \infty}f_{n+1} = {3 \over 7}\lim_{n \to \infty}f_n + 8 \\ F = {3 \over 7}F + 8 \\ F = 14$$ The problem is with if part. There are several ways to prove that $f_n$ converges for any $f_0$, but the simplest would be to take $$g_n = f_n-14, g_0 = f_0-14 = -28 \\ g_{n+1}+14 = {3 \over 7}(g_n+14) + 8 \\ g_{n+1} = {3 \over 7}g_n = ({3 \over 7})^2g_{n-1} = ... = ({3 \over 7})^{n+1}g_0$$ Thus, $\exists \lim_{n \to \infty}g_n = 0$ and $\exists \lim_{n \to \infty}f_n = 14$.

Answer 2

Note that $\lim_{n\to \infty} f_{n+1}=\lim_{n\to \infty} f_n$ if the limit exists.