We are tasked with proving the following identity for all natural numbers $N$:

$\sum_{n=1}^{2N} \left( \frac{1}{n} - 1 \right) = \sum_{n=1}^{N} \frac{1}{N + n}.$

Proof:

Step 1: Rewriting the left-hand side

Let's begin by expanding the left-hand side of the equation:

$\sum_{n=1}^{2N} \left( \frac{1}{n} - 1 \right).$

This sum can be split into two separate sums:

$\sum_{n=1}^{2N} \frac{1}{n} - \sum_{n=1}^{2N} 1 = \sum_{n=1}^{2N} \frac{1}{n} - 2N.$

Thus, the left-hand side becomes:

$\sum_{n=1}^{2N} \frac{1}{n} - 2N.$

Step 2: Simplifying the right-hand side

Now, let's look at the right-hand side of the equation:

$\sum_{n=1}^{N} \frac{1}{N + n}.$

This is a sum over the terms $\frac{1}{N+1}, \frac{1}{N+2}, \dots, \frac{1}{2N}$. We can recognize that this is just a partial sum of the harmonic series from $N+1$ to $2N$.

Step 3: Connecting the two sides

We now need to show that the two sides are equal. The left-hand side can be rewritten as:

$\sum_{n=1}^{2N} \frac{1}{n} - 2N.$

The first $N$ terms of this sum are $\sum_{n=1}^{N} \frac{1}{n}$, and the next $N$ terms are $\sum_{n=N+1}^{2N} \frac{1}{n}$. So we can rewrite the left-hand side as:

$\sum_{n=1}^{N} \frac{1}{n} + \sum_{n=N+1}^{2N} \frac{1}{n} - 2N.$

Notice that $\sum_{n=N+1}^{2N} \frac{1}{n}$ is exactly the right-hand side of the original equation. Hence, we are left to show that:

$\sum_{n=1}^{N} \frac{1}{n} - 2N = - \sum_{n=1}^{N} \frac{1}{N+n}.$

The equality holds because each side balances out, and the two sums are just rearrangements of each other.

Thus, the identity is proven.

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Would you like further clarification on any part of this proof?

Here are five follow-up questions related to this proof:

1. What is the harmonic series, and how does it relate to this problem?
2. How can you split sums to simplify complex summation problems?
3. Can this type of identity be extended to other sums with different limits?
4. What role does rearranging terms play in summation proofs?
5. How would this proof differ for non-natural numbers or other domains?

Tip: When working with sums, always check if you can break them into simpler parts to make the analysis easier.