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Considering the following recurrence: $\left(E\right):\:\:a_n=n\left(n+1\right)a_{n-1}+f\left(n\right),\:n\in \:\mathbb{N},\:n\ge \:2$

$a_n=\left[n!\right]\left[\left(n+1\right)!\right]+\frac{1}{n},\:\forall n\:\ge 1\:\:$ it's a solution for $\left(E\right):$

Find $f\left(n\right)$ and a general solution for $\left(E\right):$

This is the first time that I have to solve this type of recurrence exercise, i don't know how i can start. Should I equal $\left[n!\right]\left[\left(n+1\right)!\right]+\frac{1}{n}=n\left(n+1\right)a_{n-1}+f\left(n\right)$ to find $f\left(n\right)$ ?

Meto ballaes
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    It's a good start. You should also write down the expression for $a_{n-1}$ and substitute it into the expression. You can then solve for $f(n)$ in terms of $n$. – Theo Bendit Oct 15 '21 at 05:06
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    Also, MathJax tip: There's no need for \left and \right for every pair of brackets/parentheses. The \left and \right commands are for when you need your parentheses to scale to match what's inside them, e.g. f\left(\frac{a}{b}\right) produces $f\left(\frac{a}{b}\right)$, whereas f(\frac{a}{b}) produces $f(\frac{a}{b})$. Also, I've never noticed until now, but when using \left and \right gratuitously, it messes with spacing. Compare f(n) with f\left(n\right), which produces $f(n)$ and $f\left(n\right)$ respectively. – Theo Bendit Oct 15 '21 at 05:11
  • You'll also find it useful to note that if $a_n$ satisfies $a_n-n(n+1)a_{n-1}=f(n)$, and if $b_n$ satisfies $b_n-n(n+1)b_{n-1}=0$, then $c_n=k.b_n+a_n$ satisfies $c_n-n(n+1)c_{n-1}=k.0+f(n)=f(n)$. So you can turn a particular $a_n$ and $b_n$ into a general solution $c_n$. – Michael Hartley Oct 15 '21 at 05:19
  • @TheoBendit Ok thanks for the tip. I have wrote those math problems with a virtual math keyboard that I have made in javascript to write faster in latex – Meto ballaes Oct 15 '21 at 11:00

1 Answers1

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The homogeneous recurrence has as solution

$$ a_h(n) = c_0 (1)_n (2)_n $$

now assuming that the particular has the form $a_p(n) = c_0(n) (1)_n (2)_n$, after substitution into the complete recurrence we obtain

$$ c_0(n)-c_0(n-1) = \frac{f(n)}{\Gamma(n+1)\Gamma(n+2)} $$

thus we obtain

$$ c_0(n) = \sum_{k=0}^{n}\frac{f(k)}{\Gamma(k+1)\Gamma(k+2)} $$

and finally

$$ a(n) = (1)_n (2)_n\left(c_0 + \sum_{k=0}^{n}\frac{f(k)}{\Gamma(k+1)\Gamma(k+2)}\right) $$

Here $(z)_n = \frac{\Gamma(z+n)}{\Gamma(z)}$ is the Pochhammer function.

Cesareo
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