Question: I have found confronting information on other websites and thus will ask my question here.

Is the following conversion legitimate?

$$\sum\_{k=1}^n\sum\_{j=1}^nk\cdot j=\sum\_{k=1}^nk\cdot\sum\_{j=1}^nj$$

I have seen someone doing something similar on [Matroids Matheplanet](http://www.matheplanet.com/default3.html?call=viewtopic.php?topic=11746&ref=https%3A%2F%2Fwww.google.de%2F). 

Thank you for your help.

Philipp

Answer: 
Since you have final sums, there is no problem. See
$$
\sum\_{k=a}^n\sum\_{j=b}^m c\_kd\_j=\sum\_{k=a}^n\left(\sum\_{j=b}^m c\_kd\_j\right)=
\sum\_{k=a}^n\left(c\_k\sum\_{j=b}^m d\_j\right)=\left(\sum\_{k=a}^nc\_k\right)\left(\sum\_{j=b}^m d\_j\right)
$$
You just use the distributive law. At the first equation I added the brackets to clarify the meaning. Next $c\_k$ is in the sum over $j$ like a constant. So you can exchange them. Finally the sum over j is independent of $k$ and you can exchange them again.
> 
> Trivia: $\Large\sum, \int, \newcommand{\derivative}[3] {\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \derivative{n}{y}{x}$ are linear operations.
> 
> 
> 

What does it mean?

$\Large f(x)$ is linear if and only if

$$\Large f(x+y)=f(x)+f(y)$$

$$\Large f(ax)=a\cdot f(x)$$

Let 

$$\large f(x)=\sum\_{j=1}^nx$$

And

$$\large a=\sum\_{k=1}^nk$$

So

$$\large f(ax)= \sum\_{j=1}^n\left(\sum\_{k=1}^nk\right)x$$