Clarification on alternative expression of sum I am not familiar with alternative expression of sum shown below,
$ d_k=\sum_{i+j+l=k}a_ib_jc_l $
How it does work?
for $k = 4 $ then,
$d_{4} = \sum_{i+j+l=4}a_ib_jc_l = ...$
How do I express it in standard summation format namely $\sum_{n=0}^{\infty}$
This is from the formula to calculate the product of three summations (power series) or $\sum_{i=0}^{\infty} a_i x^i \cdot \sum_{j=0}^{\infty} b_j x^j \cdot \sum_{l=0}^{\infty} c_l x^l = \sum_{k=0}^{\infty} d_kx^k$
 A: First, it is assumed that the indices are non-negative.
Second step, it means that the sum is to be taken for the triples $(i,j,l)$ for which $i+j+l=k |0 \le i,j,l$. These are the "weak" compositions of $k$ into three parts
So for e.g. $k=4$ you have to sum $a_0b_0c_4+\cdots+a_0b_1c_3+\cdots$
A: You can get
explicit indices like this
(assuming that
the lower index of summation
is $0$):
$\begin{array}\\
d_k
&=\sum_{i+j+l=k}a_ib_jc_l\\
&=\sum_{i=0}^{k}
a_i \sum_{j=0}^{k-i}b_jc_{k-i-j}\\
\end{array}
$
If the summation
is over an inequality,
you get one more level
of summation:
$\begin{array}\\
e_k
&=\sum_{0 \le i+j+l \le k}a_ib_jc_l\\
&=\sum_{i=0}^{k}
a_i \sum_{j=0}^{k-i}b_j\sum_{l=0}^{k-i-j}c_{l}\\
\end{array}
$
Work out for yourself
what these are
if the lower index of summation
is $1$ instead of $0$.
A: The representation of $d_k$ as triple sum $\sum_{i+j+l=k}a_ib_jc_l$ can be derived by applying  Cauchy Series multiplication  twice.

We        obtain
\begin{align*}
\left(\sum_{i=0}^\infty\right.&\left.  a_i   x^i\right)\left(\sum_{j=0}^\infty   b_j x^j\right)\left(\sum_{j=0}^\infty   b_j x^j\right)\\
&=\left(\sum_{n=0}^\infty\left(\sum_{{i+j=n}\atop{i,j\geq 0}}a_ib_j\right)x^n\right)\left(\sum_{j=0}^\infty   b_j x^j\right)\tag{1}\\
&=\sum_{k=0}^\infty\left(\sum_{{n+l=k}\atop{n,l\geq 0}}\left(\sum_{{i+j=n}\atop{i,j\geq 0}}\right)c_l\right)x^k\\
&=\sum_{k=0}^\infty\left(\color{blue}{\sum_{{i+j+l=k}\atop{i,j,l}}a_ib_jc_l}\right)x^k\tag{2}\\
\end{align*}

From (1) we also obtain

\begin{align*}
\left(\sum_{n=0}^\infty\right.&\left.\left(\sum_{{i+j=n}\atop{i,j\geq 0}}a_ib_j\right)x^n\right)\left(\sum_{j=0}^\infty   b_j x^j\right)\\
&=\left(\sum_{n=0}^\infty\left(\sum_{i=0}^na_ib_{n-i}\right)x^n\right)\left(\sum_{l=0}^\infty c_lx^l\right)\\
&=\sum_{k=0}^\infty\left(\sum_{{n+l=k}\atop{n,l\geq 0}}\left(\sum_{i=0}^na_ib_{n-i}\right)c_l\right)x^l\\
&=\sum_{k=0}^\infty\left(\color{blue}{\sum_{n=0}^k\sum_{i=0}^na_ib_{n-i}c_{k-n}}\right)x^k\tag{3}
\end{align*}
  Comparing coefficients of equal powers of $x$ in (2) and (3) shows equality of the sums.

