I've reduced a large homework problem to the following smaller problem.

Let $P = \sum_{i=0}^\infty a_i X^i$ denote a formal power series over a field. Assume $a_0 \neq 0$, and define $Q = \sum_{i=0}^\infty b_i X^i$ by asserting that $$b_0 = \frac{1}{a_0}$$ and that for all $n \geq 1$ it holds that

$$b_n = \frac{-1}{a_0}\sum_{i=1}^n a_i b_{n-i}.$$

The problem is to show that $PQ=1$.

What I've done so far.

For all polynomials R, write $X^n \cdot R$ for the coefficient of $X^n$ in $R$.

It is easy to see that $X^0 \cdot PQ = 1.$ Thus, it remains to show that for all $n \geq 1$ it holds that $X^n \cdot PQ = 0$. How might I proceed from here? I've tried everything and nothing seems to work....


We have (provided appropriate convergence holds and we are allowed to swap summations) $$PQ = \left(\sum_{i=0}^\infty a_i x^i \right) \cdot \left(\sum_{j=0}^{\infty} b_j x^j\right) = \sum_{i,j=0}^{\infty} a_i b_j x^{i+j} \tag{$\star$}$$ Now if we let $i+j = m$, the above can be rewritten as $$PQ = \sum_{m=0}^{\infty} \sum_{k=0}^m a_k b_{m-k} x^m = \sum_{m=0}^{\infty} \left(\underbrace{\sum_{k=0}^m a_k b_{m-k}}_{0 \text{ for }m \neq 0} \right) x^m = a_0 b_0 = 1$$

  • $\begingroup$ Yes but the question is, how do you show that its $0$ for $m \neq 0$? $\endgroup$ Apr 5 '13 at 4:49
  • $\begingroup$ @user18921 Isn't that given in the problem? For $n \neq 0$, $$b_n = - \dfrac1{a_0} \sum_{i=1}^n a_i b_{n-i} \implies a_0 b_n = - \sum_{i=1}^n a_i b_{n-i} \implies \sum_{i=0}^n a_i b_{n-i} = 0$$ $\endgroup$
    – user17762
    Apr 5 '13 at 4:50
  • $\begingroup$ Oh wow. I can't believe I didn't see that. $\endgroup$ Apr 5 '13 at 4:52

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