Prove that $X_n=(-5)^nX_0+a_n$ where $a_n=-5a_{n-1}+\frac{1}{n}, a_0=0$

Question

Let $X_n=-5X_{n-1}+\frac{1}{n}$ for some initial value $X_0$. Prove that $X_n=(-5)^nX_0+a_n$ where $a_n=-5a_{n-1}+\frac{1}{n}, a_0=0$. Determine the condition of the calculation task $ X_{20} $ due to the disturbance $ X_0 $. Is this a good condition?

When it comes to proof of recurrence, I did not manage to come to a solution, and I don't completely understand the second part of the task.

Answer 1

This you can do with induction: By induction hypothetis we have:$$X_n=(-5)^nX_0+a_n$$ so

$$X_{n+1} = (-5)X_n +{1\over n+1}= (-5)((-5)^nX_0+a_n)+ {1\over n+1} $$ $$=(-5)^{n+1}X_0 \underbrace{-5a_n+{1\over n+1}}_{a_{n+1}} =(-5)^{n+1}X_0 +a_{n+1} $$

Answer 2

Let $ n\geq 1 $, we have : \begin{aligned} \left(\forall k\in\left[\!\left[0,n-1\right]\!\right]\right),\ X_{n-k}&=-5X_{n-k-1}+\frac{1}{n-k}\\ \iff \ \ \ \ \ \ \ \ \left(\forall k\in\left[\!\left[0,n-1\right]\!\right]\right),\ \left(-5\right)^{k}X_{n-k}&=\left(-5\right)^{k+1}X_{n-k-1}+\frac{\left(-5\right)^{k}}{n-k} \\ \Longrightarrow \sum_{k=0}^{n-1}{\left(\left(-5\right)^{k}X_{n-k}-\left(-5\right)^{k+1}X_{n-k-1}\right)}&=\sum_{k=0}^{n-1}{\frac{\left(-5\right)^{k}}{n-k}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ X_{n}-\left(-5\right)^{n}X_{0}&=\sum_{k=1}^{n}{\frac{\left(-5\right)^{n-k}}{k}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ X_{n}&=\left(-5\right)^{n}X_{0}+\sum_{k=1}^{n}{\frac{\left(-5\right)^{n-k}}{k}}\end{aligned}

Meaning, your $ a_{n} $ would be $ \sum\limits_{k=1}^{n}{\frac{\left(-5\right)^{n-k}}{k}} \cdot $