Double summation manipulation

Question

How to prove the equivalence between these two double sums? Probably it is a stupid task, but I cannot solve it.
Let $k\in\mathbb{N}$, prove that for every $\alpha,\beta,s_i\in\mathbb{R}$ it holds $$ \sum_{j=0}^{k-1}\sum_{i=0}^j\dfrac{(-1)^is_{2j}\alpha^{2j-2i}\beta^{2i+1}}{(2i+1)!(2j-2i)!}=\sum_{j=0}^{k-1}\sum_{i=0}^{k-j-1}\dfrac{(-1)^{j}s_{2i+2j}\alpha^{2i}\beta^{2j+1}}{(2j+1)!(2i)!}. $$ Maybe, beside this, there is the property of double summation $\sum_{j=0}^n\sum_{i=0}^ja_ib_{j-i}=\sum_{j=0}^{n}a_j\sum_{i=0}^{n-j}b_i$, but I cannot solve the problem.

Answer 1

Your approach is fine. Here is an answer with some additional information. We want to show the following identity is valid: \begin{align*} \sum_{j=0}^{k-1}\sum_{i=0}^j\dfrac{(-1)^is_{2j}\alpha^{2j-2i}\beta^{2i+1}}{(2i+1)!(2j-2i)!}=\sum_{j=0}^{k-1}\sum_{i=0}^{k-j-1}\dfrac{(-1)^{j}s_{2i+2j}\alpha^{2i}\beta^{2j+1}}{(2j+1)!(2i)!}\tag{1} \end{align*}

We start with the right-hand side of (1) and obtain \begin{align*} \color{blue}{\sum_{j=0}^{k-1}}&\color{blue}{\sum_{i=0}^{k-j-1} \frac{(-1)^js_{2i+2j}\alpha^{2i}\beta^{2j+1}}{(2j+1)!(2i)!}}\\ &=\sum_{i=0}^{k-1}\sum_{j=0}^{k-i-1}\frac{(-1)^is_{2i+2j}\alpha^{2j}\beta^{2i+1}}{(2i+1)!(2j)!}\tag{2}\\ &=\sum_{i=0}^{k-1}\sum_{j=i}^{k-1}\frac{(-1)^is_{2j}\alpha^{2j-2i}\beta^{2i+1}}{(2i+1)!(2j-2i)!}\tag{3}\\ &=\sum_{0\leq i\leq j\leq k-1}\frac{(-1)^is_{2j}\alpha^{2j-2i}\beta^{2i+1}}{(2i+1)!(2j-2i)!}\tag{4}\\ &\,\,\color{blue}{=\sum_{j=0}^{k-1}\sum_{i=0}^{j}\frac{(-1)^is_{2j}\alpha^{2j-2i}\beta^{2i+1}}{(2i+1)!(2j-2i)!}}\tag{5}\\ \end{align*} and the claim (1) follows.

Comment:

• In (2) we exchange the names of the indices $i\leftrightarrow j$.

• In (3) we shift the index of the inner sum to start with $j=i$.

• In (4) we write the index region more conveniently as preparation for the next step.

• In (5) we exchange inner and outer sum.

Answer 2

Probably this is the way to prove it, even if I am not sure about some details.
Let us manipulate the r.h.s. of the equality by substituting $i\to i-j$ and then switching the indexes: $$ \begin{split} \sum_{j=0}^{k-1}\sum_{i=0}^{k-j-1}\dfrac{(-1)^js_{2i+2j}\alpha^{2i}\beta^{2j+1}}{(2j+1)!(2i)!}&=\sum_{j=0}^{k-1}\sum_{i=j}^{k-1}\dfrac{(-1)^js_{2i}\alpha^{2i-2j}\beta^{2j+1}}{(2j+1)!(2i-2j)!}=\sum_{i=0}^{k-1}\sum_{j=i}^{k-1}\dfrac{(-1)^is_{2j}\alpha^{2j-2i}\beta^{2i+1}}{(2i+1)!(2j-2i)!}\\ &=\sum_{j=0}^{k-1}\sum_{i=0}^j\dfrac{(-1)^is_{2j}\alpha^{2j-2i}\beta^{2i+1}}{(2i+1)!(2j-2i)!}, \end{split} $$ where last equality follows since if $i>j$, we assume conventionally $(2j-2i)!=\infty$.
This last statement does not convince me totally but I think it works.