# Algebraic manipulation of product of two summations

#### CKKOY

If

$F_1(x)=\sum_{n\ge 0}a_nx^n$, $F_2(x)=\sum_{n\ge 0}b_nx^n$,

why does the following hold?

\begin{align*} F_1(x)F_2'(x)&=\sum_{n\ge 0}a_nx^n\sum_{n\ge 0}nb_nx^{n-1}\\ &=\sum_{n\ge 0}a_nx^n\sum_{n\ge 0}(n+1)b_{n+1}x^n\\ &=\sum_{n\ge 0}\sum_{k=0}^na_k(n-k+1)b_{n-k+1}x^n \end{align*} More specifically, how does one move from the second to the third line of the derivative-related equation?

#### v8archie

Math Team
It's just a term by term treatment of the product, looking at increasing powers of $x$.

1 person

#### mathman

Forum Staff
To save writing, let $$\displaystyle c_n=(n+1)b_{n+1}$$.

The second line is then $$\displaystyle \sum_{n\ge 0}\sum_{k\ge 0}a_nc_kx^{n+k}$$.
Now let j=n+k so the double sum becomes $$\displaystyle \sum_{n\ge 0}\sum_{j\ge n}a_nc_{j-n}x^j$$.

Finally switch the order of summations. $$\displaystyle \sum_{j\ge 0}x^j\sum_{n=0}^j a_nc_{j-n}$$.

1 person