rewriting product of power series 
According to
  $$\Lambda(\tau;q)=B_0(\tau)+\sum_{i=1}^\infty B_i (\tau) q^i$$
we define
  $$[\Lambda(\tau;q)-B_0(\tau)]^m=\bigg[ \sum_{i=1}^\infty B_i (\tau) q^i \bigg]^m=\sum_{n=m}^{\infty}\mu_{m,n}(\tau)q^n,$$
  with the definition
  $$\mu_{1,n}(\tau)=B_n(\tau), \qquad n\geq1.$$
  Then, it holds
  $$[\Lambda(\tau;q)-B_0(\tau)]^{m+1}=\sum_{n=m+1}^{\infty}\mu_{m+1,n}(\tau)q^n,$$
  $$=\bigg[ \sum_{n=m}^{\infty}\mu_{m,n}(\tau)q^n \bigg] \bigg[ \sum_{i=1}^\infty B_i (\tau) q^i \bigg] = \sum_{n=m+1}^{\infty} q^n \bigg[ \sum_{i=m}^{n-1}\mu_{m,i}(\tau) B_{n-i}(\tau)\bigg]$$
  which gives the recursion formula
  $$\mu_{m+1,n}(\tau)=\sum_{i=m}^{n-1}\mu_{m,i}(\tau) B_{n-i}(\tau)$$

I don't understand how do we get 
$$\bigg[ \sum_{n=m}^{\infty}\mu_{m,n}(\tau)q^n \bigg] \bigg[ \sum_{i=1}^\infty B_i (\tau) q^i \bigg] = \sum_{n=m+1}^{\infty} q^n \bigg[ \sum_{i=m}^{n-1}\mu_{m,i}(\tau) B_{n-i}(\tau)\bigg]$$
The expression in left hand side is a product of two power series. How can it be rewritten as the right hand side? I think there are some omitted steps and I wish to know them. Can anyone help me?
Thanks.
 A: The calculation is an application of the Cauchy product formula of two power series:
\begin{align*}
\left(\sum_{j=m}^\infty a_j q^j\right)\left(\sum_{k=n}^\infty b_k q^k\right)
&=\sum_{l=m+n}^\infty\left(\sum_{{j+k=l}\atop{j\geq m;k\geq n}}a_j b_k\right)q^l\\
&=\sum_{l=m+n}^\infty\left(\sum_{j=m}^{l-n}a_j b_{l-j}\right)q^{l}
\end{align*}

We obtain 
  \begin{align*}
[\Lambda(\tau;q)-B_0(\tau)]^{m+1}&=\left[ \sum_{i=1}^\infty B_i (\tau) q^i \right]^{m+1}\\
&=\left[ \sum_{i=1}^\infty B_i (\tau) q^i \right]^{m}\left[ \sum_{i=1}^\infty B_i (\tau) q^i \right]\\
&=\left[\sum_{n=m}^{\infty}\mu_{m,n}(\tau)q^n\right]\left[\sum_{i=1}^\infty B_i (\tau) q^i \right]\\
&=\sum_{r={m+1}}^\infty\left[\sum_{{n+i=r}\atop{n\geq m; i\geq 1}}\mu_{m,n}(\tau)B_i (\tau)\right]q^r\tag{1}\\
&=\sum_{r={m+1}}^\infty\left[\sum_{n=m}^{r-1}\mu_{m,n}(\tau)B_{r-n}(\tau)\right]q^r\tag{2}\\
&=\sum_{n={m+1}}^\infty\left[\sum_{i=m}^{n-1}\mu_{m,i}(\tau)B_{n-i} (\tau)\right]q^n\tag{3}
\end{align*}

Comment:


*

*In (1) we apply the Cauchy product formula

*In (2) we replace $i$ with $r-n$ respecting the condtion: $n+i=r$. The upper limit of the inner sum is $r-1$ since $n+i=r$ and $i\geq 1$.

*In (3) we rename $n\rightarrow i$ and $r \rightarrow n$
