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I'm trying to solve the following recurrence relation, but I have a problem with the factorial part. I would like to evaluate its particular solution. I would like also to suggest a textbook for solving these kinds of questions. We have:

$$a_n = \sqrt{n(n+1)}a_{n-1} + n!(n+1)^{\frac{3}{2}}$$

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    $\begingroup$ Hi @AhmedSaleh, I've formated your equation with latex so that possible future edits become easier. I encourage you to learn the basics of typing math with latex so that you can get the best out of Math.SE! Good luck! $\endgroup$
    – Gold
    May 16, 2013 at 23:13

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OK, here's my solution. Note you didn't give any boundary values. Your equation can be rewritten as $$ \frac{a_n}{n!}=\frac{a_{n-1}\sqrt{n(n+1)}}{n!}+(n+1)^{\frac{3}{2}} $$ Denote $b_n=\frac{a_n}{n!}$: $$ b_n=b_{n-1}\sqrt{\frac{n+1}{n}}+(n+1)^{\frac{3}{2}}\\ \frac{b_n}{\sqrt{n+1}}=\frac{b_{n-1}}{\sqrt{n}}+(n+1)^3 $$ Denote $s_n=\frac{b_{n}}{\sqrt{n+1}}$: $$ s_n=s_{n-1}+(n+1)^3=\ldots=s_0+S_n $$ here $$ S_n=\sum_{k=1}^{n+1}k^3=O(n^4) $$ I don't give the calculation of this expression, as you can easily find it by methods such as perturbation, induction, etc. Hence your expression is $$ a_n=n!\sqrt{n+1}(a_0+S_n) $$

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  • $\begingroup$ your proof is perfect except that the sum starts from k=2 instead of k=1. $\endgroup$
    – toufik_kh.17
    Dec 4, 2013 at 10:18

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