$a_0 = 1 \\ a_{n+1} = (n+1)a_n + 1 $
Could you help me solve this? And maybe someone know good source explaining how to solve recurrence relations?
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Hint: Try to find and solve the recursion for $b_n=a_n/n!$. – Did Oct 19 '13 at 11:47
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Hint: you can use generating function technique. – Mhenni Benghorbal Oct 19 '13 at 12:07
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as @Did said in the comments we can put
$$ n!b_n = a_n \Rightarrow (n+1)!b_{n+1} = (n+1)!b_n + 1 $$
$$ \Rightarrow (b_{n+1} - b_n)(n+1)! = 1 \Rightarrow b_{n+1} - b_n = \frac{1}{(n+1)!} $$
you can do the following because you'll get a telescoping sum $$ \sum_{n=0}^k (b_{n+1} - b_n) = \sum_{n=1}^{k+1} \frac{1}{n!} \Rightarrow b_{k+1} - b_0 = \sum_{n=1}^{k+1} \frac{1}{n!} $$
$$ a_0 = 1 = 0!b_0 = b_0 \Rightarrow b_{k+1} = 1 + \sum_{n=1}^{k+1} \frac{1}{n!} \Rightarrow b_n = 1 + \sum_{k=1}^{n} \frac{1}{k!} $$
$$ \Rightarrow a_n= n! \left(1 + \sum_{k=1}^{n} \frac{1}{k!} \right) $$
what'sup
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It's always a good idea to compute the first few terms of a sequence and check them with the OEIS. In this case you get A000522.
Barry Cipra
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