Why is:$$\sum_{k=1}^\infty \sum_{n=k}^\infty \frac{1}{n(n+1)} k\, a_k =\sum_{1\le k\le n}\frac{1}{n(n+1)} k\, a_k=\sum_{n=1}^\infty \frac{1}{n+1}\; \frac{1}{n}\sum_{k=1}^n k\, a_k$$
What does the middle term mean and why is it equal to the first? I thought $\sum_{1\le k\le n}=\sum \limits_{k=1}^{n}$.
On
That would be my first instinct as well, but whoever wrote this line must have intended it to mean a sum over two variables $k$ and $n$. In my view, I imagine the author intended the middle expression to help people understand why the first and last expressions are equal (in other words, to help explain the exchange of summations); but you can simply ignore the middle expression if you already believe that the outer expressions are equal.
On
The symbol $\sum_{1\le k\le n}$ is a bit unusual, but in the present context it means that we take the sum over thes set of pairs of integers $P = \{(k,n) \in \mathbb N \times \mathbb N \mid 1 \le k \le n \}$.
But how is the value of a "series" of the form $\sum_{(k.n) \in P} u_{k,n}$ defined? We say that $\sum_{(k,n) \in P} u_{k,n} = u$ if for all $\epsilon > 0$ there exists a finite subset $P_\epsilon \subset P$ such that for all finite subsets $P' \subset P$ with $P_\epsilon \subset P'$ $$\left\lvert \sum_{(k,n) \in P'} u_{k,n} - u \right\rvert <\epsilon .$$
Of course it requires a proof that the first equation is true (we have to compare various "infinite sums"). However, it is obvious that each single summand $\frac{1}{n(n+1)} k a_k$ oocurs on both sides of the equation. The issue is to prove that the limit of the RHS exists and equals the LHS. I shall not do that here.
$$ \sum_{k=1}^\infty \sum_{n=k}^\infty = \sum_{n=1}^\infty \sum_{k=1}^n $$ where the intermediate notation is a lazy notation where actually a double sum ought to be used. This is an obvious resummation in a infinite triangular region spanned by $k$ and $n$, see e.g. the article by Choi "Notes on formal manipulations of double series" in https://doi.org/10.4134/CKMS.2003.18.4.781 .