A sequence $(b_n)_n$ is given as $b_0 = 1$ and for every $n \in \mathbb N$
$$\sum_{k=0}^n\binom{n+1}{k}b_k=0\tag{1}$$
The task is to find its exponential generating function $f(x)$
From $(1)$ we get
$$\sum_{k=0}^n\binom{n}{k}b_k=\sum_{k=0}^{n-1}\binom{n}{k}b_k+b_n=b_n$$
$$f(x)=\sum_{n=0}^\infty\frac{b_n}{n!}x^n$$
$$f(x)e^x=\left(\sum_{n=0}^\infty\frac{b_n}{n!}x^n\right)\left(\sum_{n=0}^\infty\frac{1}{n!}x^n\right)=\sum_{n=0}^\infty\frac{1}{n!}x^n\left(\sum_{k=0}^n\binom{n}{k}b_k\right)=\sum_{n=0}^\infty\frac{b_n}{n!}x^n=f(x)$$
$$f(x)(e^x-1)=0$$
Now i am not sure what to do or if i made a mistake.
Best Answer
The mistake is very subtle; equation $(1)$ only holds for $n\neq 0 $, which implies that $$\sum_{k=0}^n\binom{n}kb_k=b_n\tag{2}$$ only holds for $n\neq 1$. See that when $n=1$, equation $(2)$ becomes $$ \sum_{k=0}^1\binom{n}kb_k=\binom10b_0+\binom11b_1=b_1+1\neq b_1 $$
The lesson is that most of the time, the LHS of $(2)$ is equal to $b_n$, but when $n=1$, the LHS is equal to $b_1+1$. Therefore, starting with $$ f(x)e^x=\sum_{n=0}^\infty \frac{x^n}{n!}\left(\sum_{k=0}^n\binom{n}kb_k\right), $$ almost all of the coefficients in parentheses can be replaced with $b_n$, but the $n=1$ term needs a correction: \begin{align} f(x)e^x &=b_0+(b_1+1)\frac{x^1}{1!}+\sum_{n=2}^\infty b_n\frac{x^n}{n!} \\&=f(x)+x. \end{align}Using this, you can now solve for $f$.