Product of two finite sums $\left(\sum_{k=0}^{n}a_k\right) \left(\sum_{k=0}^{n}b_k\right)$

Question

What is the product of the following summation with itself:

$$\left(\sum_{k=2}^{n}k(k-1)a_{k}t^{k-2} \right) \left(\sum_{k=2}^{n}k(k-1)a_{k}t^{k-2}\right) $$

Is the above equal to the double summation below and if not, what is the correct proof or derivation? $$\sum_{k=2}^{2n}\left(\sum_{i=2}^{k}k(k-1)a_{i}a_{k-i}\right)t^{k-2}$$

I based my answer on the following rule for the product of two summations. However, it is also not clear how they derived this (i.e. where does the upper bound $2n$ come from in the first summation and the total second summation). What is the exact derivation of this step by step? $$\left(\sum_{k=0}^{n}a_{k} \right) \left(\sum_{k=0}^{n}b_{k}\right) = \sum_{k=0}^{2n}\left(\sum_{i=\max(0,k-n)}^{\min(n,k)}a_{i}b_{k-i}\right)$$

Answer 1

We consider the Cauchy product of two polynomials in $z$.

We obtain \begin{align*} \color{blue}{\left(\sum_{k=0}^n\right.}&\color{blue}{a_kz^k\Bigg)\left(\sum_{l=0}^nb_lz^l\right)}\\ &=\sum_{q=0}^{2n}\left(\sum_{{k+l=q}\atop{0\leq k,l\leq n}}a_kb_l\right)z^q\tag{1}\\ &=\sum_{q=0}^{2n}\left(\sum_{{{k=0}\atop{0\leq k\leq n}}\atop{0\leq q-k\leq n}}^qa_kb_{q-k}\right)z^q\tag{2}\\ &=\sum_{q=0}^{n}\left(\sum_{k=0}^qa_kb_{q-k}\right)z^q +\sum_{q=n+1}^{2n}\left(\sum_{k=q-n}^na_kb_{q-k}\right)z^q\tag{3}\\ &\,\,\color{blue}{=\sum_{q=0}^{2n}\left(\sum_{k=\max\{0,q-n\}}^{\min\{q,n\}}a_kb_{q-k}\right)z^q}\tag{4} \end{align*}

Comment:

• In (1) we see the greatest power of $z$ is $2n$ explaining the upper limit of the outer sum.

• In (2) we eliminate the index $l$ using $l=q-k$, but we still have to respect the conditions on $k$ and $l=q-k$.

• In (3) we split the sum and respect explicitly the conditions $0\leq k\leq n$ and $0\leq q-k\leq n$.

• In (4) we merge the sums from (3).

  • Note the outer sum in (4) has $2n+1$ summands with index $q$ ranging from $0$ to $2n$. This coincides with the $n+1$ summands from the left-hand outer sum of (3) with index $q$ ranging from $0\leq q\leq n$ and the $n$ summands from $n+1$ to $2n$ from the right-hand outer sum of (3) with index $q$ ranging from $n+1$ to $2n$.

  • The summand $a_kb_{q-k}$ is the same in both inner sums of (3) which is also taken in (4).

  • Comparing lower and upper limit of the inner sums in (3) with $\max\{0,q-n\}\leq k\leq \min\{q,n\}$ and respecting the lower and upper limits of the outer index $q$ shows equality.

Evaluating (4) at $z=1$ we obtain the relation \begin{align*} &\color{blue}{\left(\sum_{k=0}^{n}a_k\right) \left(\sum_{k=0}^{n}b_k\right)}\color{blue}{= \sum_{q=0}^{2n}\left(\sum_{k=\max\{0,q-n\}}^{\min\{q,n\}}a_kb_{q-k}\right)} \end{align*} in accordance with OPs claim.

Answer 2

$$\left(\sum_{k=0}^{n}a_{k} \right) \left(\sum_{k=0}^{n}b_{k}\right) = \sum_{k=0}^{2n}\left(\sum_{i=0}^{k}a_{i}b_{k-i}\right)$$

You can think in the following way, first we list the LHS

$$(a_0+a_1+\dots+a_n)(b_0+b_1+\dots+b_n)$$

If we expand it, we will get terms like $a_i b_j$, now we can group these terms by the sum of index $i+j$, for example,

$$\begin{align}i+j=0&\Longleftrightarrow i=0, j=0\\ \\ i+j=1&\Longleftrightarrow (i=0,j=1)\lor (i=1,j=0)\\ \\ i+j=2&\Longleftrightarrow (i=0,j=2)\lor (i=2,j=0)\lor (i=1,j=1)\\ \\ \dots\end{align}$$

If we set the index sum be $i+j=k$, we get the group for product with the sum of index equals $k$, which is

$$\sum_{i=0}^{k}a_{i}b_{k-i}$$

Next, we can see the $k$ goes from $0+0=0$ until $n+n=2n$, therefore

$$\left(\sum_{k=0}^{n}a_{k} \right) \left(\sum_{k=0}^{n}b_{k}\right) = \sum_{k=0}^{2n}\left(\sum_{i=0}^{k}a_{i}b_{k-i}\right)$$

Can you proceed to verify your result? (your result in the OP is incorrect.)

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Remarks:

For those terms $a_k, b_k$, where $k>n$, we set them to be zero.