Simple Cases of hadamard product

Question

I have a simple recursive formula $a_{n+1} = (1+ 1/n) a_n +1$ and try to find the generating function of ${a^2_n}$. A classic approach is using Hadamard product of the generating function of A(x) with itself. However it leads to complex integrals. Is there any algebraic way to find ${a^2_n}$ recursively and then compute its generating functions directly as for the Fibonacci sequence. I tried but can not find a way. Any help is appreciated.

Michael Rozenberg · Accepted Answer · 2017-07-27T19:33:09.377

1

It's $$\frac{a_{n+1}}{n+1}=\frac{a_n}{n}+\frac{1}{n+1}$$ Hence, by the telescopic sum we obtain: $$\frac{a_n}{n}=a_1+H_n-1$$ or $$a_n^2=\left(na_1+n\left(H_n-1\right)\right)^2,$$ where $H_n=1+\frac{1}{2}+...+\frac{1}{n}$.

edited Jul 27 '17 at 19:33

answered Jul 27 '17 at 19:25

Michael Rozenberg

194,933

How it relates to a recursion for $a^2_n$, Please advise. – Rasoul Jul 27 '17 at 19:28
@Rasoul I fixed. See now, please. – Michael Rozenberg Jul 27 '17 at 19:34
Thanks, what happens if $a_{n+1} = (1+ {a_0}/n) a_n +1$, Is there any general approach? – Rasoul Jul 27 '17 at 20:26
@Rasoul It seems hard. I think. – Michael Rozenberg Jul 27 '17 at 20:38

Simple Cases of hadamard product

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