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The problem I am trying to do asks what $\sum{a_n}_{n=1}^{\infty}$ is if $a_{n}$ is defined as

$n(n-1)a_n = (n-1)(n-2)a_{n-1} - (n-3)a_{n-2}$ where $a_0 = a_1 = 1$

I plugged in the first few terms and found that $a_n = \frac{1}{n!}$ (which would make the sum $e$ based on the Maclaurin Series) but I'm not sure how to prove this fact from the recursive function given. Are there any general methods to use for these types of problems?

Abhi V
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  • The approach is induction. See https://math.stackexchange.com/questions/19485/dominoes-and-induction-or-how-does-induction-work/19488#19488 for an introduction. – Ross Millikan May 11 '18 at 03:00
  • You can also look up generating functions. They are great for solving recurrences. – Manan May 11 '18 at 03:16

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You can rewrite the recursive function like this: $$(n-1)(na_n - a_{n-1}) = (n-3)((n-1)a_{n-1} - a_{n-2}).$$

Setting $b_{n-1} = na_n - a_{n-1}$, then $(n-1)b_{n-1} = (n-3)b_{n-2}$ and $b_0 = 0$. Now, you have $b_n = 0$, $\forall n$. Then $a_n = \frac{1}{n}a_{n-1}.$ Form here you have $a_n = \frac{1}{n!}.$

GAVD
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