We place ourselves in the ring of formal power series.
Let $(u_k)$ be a family of natural integers.
How do I prove rigorously the following equality ?
$$\sum_{n\in N}\sum_{k=0}^nu_kT^n=\sum_{k\in N}\sum_{l\in N}u_kT^{k+l}$$
This is taken from a correction of an exam.
I am guessing the equality is not true because
$$\sum_{n\in N}\sum_{k=0}^nu_kT^n=\sum_{n\in N}(T^n\sum_{k=0}^nu_k)$$ but
$$\sum_{k\in N}\sum_{l\in N}u_kT^{k+l}=\sum_{k\in N}(T^k\sum_{l\in N}u_kT^{l})$$
They will be equal if the coefficient of $T^n$ is the same for all $n$.
So on the one hand, on the left side, the coefficient of $T^n$ is $\displaystyle \sum_{k = 0}^n u_k$.
On the other hand, the right hand side is $ \displaystyle \sum_l \sum_k u_k T^{l+k}$.What we're interested in is the case when $l+k = n$, so that we compute and compare the $T^n$th coefficient. For each $k$ such there is is a solution to $l+k=n$, we will get the term $u_k$ in the sum of the coefficients. Fortunately, the solutions to $l+k=n$ are easy to see, and start $l=n,k=0$, then $l=n-1, k = 1$, $\ldots$, $l=0, k=n$. So for every $k$ with $0 \leq k \leq n$, we include $u_k$ in the coefficient, and moreover these are all the coefficients. Thus the coefficient of $T^n$ is exactly $\displaystyle \sum_{k=0}^n u_k$.
Written more succinctly,
$$\sum_l \sum_k u_k T^{l+k} = \sum_n \sum_{l+k = n} u_k T^n = \sum_n \sum_{k = 0}^n u_k T^n,$$
where I collected the terms $l+k = n$ and summed over all possible $l,k \in \mathbb{N}$ s.t. $l+k=n$, and then realized that $l$ appears nowhere so there's no need to include it in the computation.
