do you know what is the method to find out the limit of this:
0 < q < 1
$$\sum_{n=0}^{\infty} (n+1)(n+2)q^n$$
The $q^n$ is ok, that is the geometric series, it is $\frac{1}{1-q}$.
But how do I do the Cauchy product? For me it seems, this is divergent...
Thank you!
$\endgroup$ 21 Answer
$\begingroup$To do this without differentiation, note that $$ \begin{eqnarray} \sum_{n=0}^{\infty}(n+1)q^n &=& 1+2q+3q^2+4q^3+\ldots \\ &=& 1+(1+1)q+(1+1+1)q^2+(1+1+1+1)q^3+\ldots \\ &=& (1+q+q^2+\ldots) + (q + q^2 + q^3+\ldots) + (q^2+q^3+q^4+\ldots)+\ldots \\ &=&(1+q+q^2+\ldots)(1+q+q^2+\ldots) \\ &=&\frac{1}{(1-q)^2}. \end{eqnarray} $$ Similarly, it follows that $$ \begin{eqnarray} \sum_{n=0}^{\infty}\frac{1}{2}(n+2)(n+1)q^n &=& 1+3q+6q^2+10q^3+15q^4+\ldots \\ &=& 1+(2+1)q+(3+2+1)q^2+(4+3+2+1)q^3+\ldots \\ &=& (1+2q+3q^2+\ldots) + (q+2q^2+3q^3+\ldots) + (q^2+2q^3+3q^4+\ldots) + \ldots \\ &=& (1+2q+3q^2+\ldots)(1+q+q^2+\ldots) \\ &=& \frac{1}{(1-q)^3}. \end{eqnarray} $$ Generally speaking, this is a convolution trick: $$ \sum_{n}a_n x^n \cdot \sum_{n}b_n x^n = \sum_{n}\left(\sum_{j=0}^{n}a_j b_{n-j}\right)x^n\equiv\sum_{n} \left(a *b\right)_n x^n, $$ and $(1 * 1)_n = n+1$, and $(1 * (1 * 1))_n = \frac{1}{2}(n+2)(n+1)$, and so on.
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