change order of double sums

I need to prove that

$$\sum_{k=1}^n \sum_{i=0}^{k-1} \sum_{j=k}^n C(i,j,n) \log(n+k) = \sum_{j=1}^n \sum_{i=0}^{j-1} \left(\sum_{k=i+1}^{j}C(i,j,n) \log(n+k)\right)$$

where $$C(i,j,n)={n \choose i}{n \choose j}\frac{(-1)^{i+j-1}}{j-i}$$

I tried the following :

Let $S(n)=\displaystyle\sum_{j=1}^n\displaystyle\sum_{i=0}^{j-1} \left(\displaystyle\sum_{k=i+1}^{j}C(i,j,n)log(n+k)\right)$ then each term of $ S (n) $ have a coefficient of the form $ log (n + l) $ with $ 1 \leq l \leq n $. From the above we can write $ S (n) $ as $S(n)=\displaystyle\sum_{l=1}^n log(n+l)a_l$. For every $ 1 \leq l \leq n $ I tried to find every $ a_l $, but i couldn't prove the equality.

Solution 1:

We can do it as follows : \begin{aligned}\sum_{k=1}^{n}{\color{blue}{\sum_{i=0}^{k-1}}{\color{red}{\sum_{j=k}^{n}}{a_{i,j,k}}}}&=\sum_{k=1}^{n}{\color{red}{\sum_{j=k}^{n}}{\color{blue}{\sum_{i=0}^{k-1}}{a_{i,j,k}}}}\\ &=\color{green}{\sum_{k=1}^{n}}{\color{red}{\sum_{j=k}^{n}}{\sum_{i=0}^{k-1}{a_{i,j,k}}}}\\ &=\color{red}{\sum_{j=1}^{n}}{\color{green}{\sum_{k=1}^{j}}{\sum_{i=0}^{k-1}{a_{i,j,k}}}}\\ &=\sum_{j=1}^{n}{\color{green}{\sum_{k=1}^{j}}{\color{blue}{\sum_{i=0}^{k-1}}{a_{i,j,k}}}}\\ \sum_{k=1}^{n}{\sum_{i=0}^{k-1}{\sum_{j=k}^{n}{a_{i,j,k}}}}&=\sum_{j=1}^{n}{\color{blue}{\sum_{i=0}^{j-1}}{\color{green}{\sum_{k=i+1}^{j}}{a_{i,j,k}}}}\end{aligned}

In the third and fifth line, we've used the formula for exchanging double sums : $$ \sum_{j=1}^{n}{\sum_{i=1}^{j}{u_{i,j}}}=\sum_{1\leq i\leq j\leq n}{a_{i,j}}=\sum_{i=1}^{n}{\sum_{j=i}^{n}{u_{i,j}}} $$

If you're not familiar with it, here is one simple proof using pure algebra : \begin{aligned}\displaystyle\sum_{i=1}^{n}{\displaystyle\sum_{j=i}^{n}{u_{i,j}}}&=\displaystyle\sum_{k=1}^{n}{\left(\displaystyle\sum_{i=1}^{k}{\displaystyle\sum_{j=i}^{k}{u_{i,j}}}-\displaystyle\sum_{i=1}^{k-1}{\displaystyle\sum_{j=i}^{k-1}{u_{i,j}}}\right)} \\ &=\displaystyle\sum_{k=1}^{n}{\displaystyle\sum_{i=1}^{k}{\left(\displaystyle\sum_{j=i}^{k}{u_{i,j}}-\displaystyle\sum_{j=i}^{k-1}{u_{i,j}}\right)}}\\ &=\displaystyle\sum_{k=1}^{n}{\displaystyle\sum_{i=1}^{k}{u_{i,k}}} \\ \displaystyle\sum_{i=1}^{n}{\displaystyle\sum_{j=i}^{n}{u_{i,j}}}&=\displaystyle\sum_{j=1}^{n}{\displaystyle\sum_{i=1}^{j}{u_{i,j}}}\end{aligned}