A commonly (though not universally) followed convention is that the scope of a summation is broken by addition or subtraction (but not multiplication). So if you write something like $$\sum_i a_i +1$$ that means $$\left(\sum_i a_i\right)+1$$ and not $$\sum_i(a_i+1).$$ Similarly, in your example, the summations would be broken up by the addition and subtraction between them, so they are not nested.
Note that in contrast, if there are multiplications between summations, then the multiplication is evaluated inside the summation and so the sums are nested. For instance, $$\sum_i a_i\sum_{j>i} b_{ij}$$ is commonly written with the meaning $$\sum_i \left(a_i \sum_{j>i} b_{ij}\right)$$ (and so the inner sum is a sum over values of $j$, not values of the pair $(i,j)$, since the $i$ is bound instead by the outer sum).
That said, these conventions are not universal and are typically not written down and taught explicitly, so you will probably encounter some deviations from them. If you are ever worried about whether the intended meaning is clear in something you're writing, it's always safe to insert parentheses to clarify the meaning (as I did in the examples above).
Welcome to MSE!
First, your idea regarding lagrange interpolation is a great one! The key insight is to look at particular coefficients in order to make your life easier.
So, let's let $a_1, \ldots a_n$ be some complex numbers. Then let's find a polynomial interpolating the points
$(a_1, 1), (a_2, 1), \ldots, (a_n, 1)$.
On the one hand, we know that this is the constant $1$ polynomial. On the other, we know it's some horrible linear combination:
$$
1 = \sum_{k=1}^n \prod_{j \neq k} \frac{x - a_j}{a_k - a_j}
$$
But now let's compare the $x^{n-1}$ coefficient of both sides! On the left hand side, we get $0$. On the right hand side, we get $\sum_{k=1}^n \prod_{j \neq k} \frac{1}{a_k - a_j}$.
Similarly, let's compare the $x^0$ coefficients! On the left hand side we get $1$, and on the right hand side we get $\sum_{k=1}^n \prod_{j \neq k} \frac{-a_j}{a_k - a_j} = \sum_{k=1}^n \prod_{j \neq k} \frac{a_j}{a_j - a_k}$.
To see why these are both true, let's separate the numerators and denominators of our polynomial:
$$
1 =
\sum_{k=1}^n
\left (
\prod_{j \neq k} \left ( \frac{1}{a_k - a_j} \right )
\prod_{j \neq k} \left ( x - a_j \right )
\right )
$$
Now, if we want the $x^{n-1}$ term, we need to get the $x^{n-1}$ term of each product. But imagine foiling out $\prod_{j \neq k} (x - a_j)$. The coefficient of $x^{n-1}$ is $1$. If we do this for each summand, we find
$$
0 x^{n-1} = \sum_{k=1}^n \prod_{j \neq k} \left ( \frac{1}{a_k - a_j} \right ) x^{n-1}
$$
which gives the claim.
Similarly, if we want the constant term, we need to get the constant term of each product. Again, if we imagine foiling out the $\prod_{j \neq k} (x - a_j)$, we find the constant term is exactly $\prod_{j \neq k} (-a_j)$. Doing this to each term of the sum shows
$$
1 =
\sum_{k=1}^n
\left (
\prod_{j \neq k} \left ( \frac{1}{a_k - a_j} \right )
\prod_{j \neq k} (- a_j)
\right )
$$
now if we recombine these products (and juggle some minus signs), we get the desired
$$
1 = \sum_{k=1}^n \prod_{j \neq k} \frac{a_j}{a_j - a_k}.
$$
Now, let's address some of your closing questions.
- Where might one begin with these problems?
Exactly where you began! Working out lots of concrete examples, conjecturing something to be true, and then trying to prove it! Your idea to look at lagrange interpolation was an excellent one, and it was only the idea to look at the polynomials one coefficient at a time that you were lacking. I'm sure if you spent another day or two with this problem you would have gotten it.
As a slightly more productive answer, you need to know what's the right level of generality to use when working out "concrete" examples. For instance, it's probably best in this problem to keep the $a_i$ as symbolic constants, rather than considering the special case of, say $a_1 = a_2 = a_3 = 2$. That said, it is fruitful to work with the special cases $n = 1, 2, 3, \ldots, 5$, say. These can all be worked out without too much hassle (especially if you use a computer algebra system, as you mention you did), and can give you some useful information.
In particular, the notion of a normal form is extremely useful, and whenever you're working with two objects that you know are equal, you should try and put them into some kind of normal form and compare the pieces separately. For this problem, that means expanding the lagrange polynomial into its normal form
$c_0 + c_1 x + c_2 x^2 + \ldots + c_{n-1} x^{n-1}$. If you had done this for some small examples, you would have doubtless seen the pattern in the coefficients that later became the proof.
In an informal way, problems that involve both addition and multiplication simultaneously are "hard". This "prototheorem" can be made precise in a number of ways (see skolem and presburger arithmetic compared with robinson arithmetic, for instance), but informally it permeates lots of questions. There are lots of open problems whose difficulty lies, in some sense, in trying to see how addition and multiplication interact. This means that problems that blend addition and multiplication are necessarily trickier than problems which only feature one of the two. I'm sure that this is also accurate to your lived experience.
Of course, I'm sure there are some tricks for dealing with these things, but I'm far from an expert, so I don't know any off the top of my head. I tend to tackle these on a fairly ad-hoc basis, and would also love to hear about a more systematic approach!
I hope this helps ^_^
Best Answer
Here's a simplification for the partial sum for you. Let $b_j = \mu a_j \in (0,1)$, and define $L_{n} := \sum_{j=1}^n \prod_{k=j+1}^n b_k$ and $M_{n} := \sum_{j=1}^n \prod_{k=j+1}^n b_{k-1}$.
Then your infinite sum is equal to $$\lim_{n\rightarrow\infty} L_{n} - \lim_{n\rightarrow\infty} \frac{b_n}{\mu}M_{n}.$$
Note that $M_n$ is just $L_n$ with $b_i \rightarrow b_{i-1}$ so you only need to calculate one of the two terms. This looks to me to be quite a bit simpler than your original form for the sum.
If you define $x := \min_{k=2,...,n}\{b_k\}$ and $y = \max_{k=2,...,n} \{b_k\}$ then using the geometric series I believe you have $$\frac{1-x^n}{1-x} \leq L_n \leq \frac{1-y^n}{1-y},$$ which may be relevant in understanding convergence of the limit.