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I have a recurrence equation and I have no idea how to solve it. $$a(n)=(1-\frac{1}{n}+\frac{X}{n})a(n-1)+\frac{2}{n}$$

$$a(1)=1$$ where X is a constant

I tried to put it into Wolfram and I got the following solution:

$$a(n)=\frac{\frac{(X+1)\Gamma(n+X)}{\Gamma(n+1)\Gamma(X+1)}-2}{X-1}$$

It is the correct solution. But Wolfram didn't give any intermediate steps so I am still puzzled how it is solved.

Any hints or comments are welcomed. Thanks in advance.

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  • $\begingroup$ I guess that this comes from Pochhammer symbols. Let $t=X-1$ and comute the very first terms to notice a pattern. $\endgroup$ – Claude Leibovici Jul 5 '18 at 8:05
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Let $\,k=X-1 \ne 0\,$, then the recurrence is $\displaystyle\,a_n=\frac{n+k}{n}a_{n-1}+\frac{2}{n}\,$.

Multiplying by $\,k\,$ and defining $\,b_n = k a_n\,$ gives $\displaystyle\,b_n=\frac{n+k}{n}b_{n-1}+\frac{2k}{n}\,$.

Adding $\,2\,$ on both sides gives $\displaystyle\,b_n+2=\frac{n+k}{n}b_{n-1}+\frac{2k}{n}+2=\frac{n+k}{n}(b_{n-1}+2)\,$.

Then $\,c_n=b_n+2\,$ telescopes to:

$$ \begin{align} c_n &= \frac{n+k}{n}c_{n-1} \\ &=\frac{(n+k)(n+k-1)}{n(n-1)} c_{n-2} \\ &\ldots \\ &= \frac{(n+k)(n+k-1)\ldots(k+1)}{n(n-1)\ldots 2} c_{1} \\ \end{align} $$

Substituting back into $\,a_n=\dfrac{c_n-2}{k}\,$ gives an expression equivalent to the posted one.

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