证明： $\displaystyle \frac{2n+1}{\sqrt{n^2+2n+\sqrt{n^2-1}}} = \sqrt{n^2+2n} - \sqrt{n^2-1}$。 由此求 $\displaystyle \sum_{n=1}^N \frac{2n+1}{\sqrt{n^2+2n}+\sqrt{n^2-1}}$ 的值。

1. Relate the new summand to the earlier expression by considering a shift in the index.

Let the term from the previous part be $u_n = \frac{2n+1}{\sqrt{n^2+2n}+\sqrt{n^2-1}}$.
Substituting $n$ with $n-1$, we get:
$u_{n-1} = \frac{2(n-1)+1}{\sqrt{(n-1)^2+2(n-1)}+\sqrt{(n-1)^2-1}}$
$u_{n-1} = \frac{2n-1}{\sqrt{n^2-2n+1+2n-2}+\sqrt{n^2-2n+1-1}}$
$u_{n-1} = \frac{2n-1}{\sqrt{n^2-1}+\sqrt{n^2-2n}}$
This exactly matches the summand in the target sum for $4(a)$.

2. Rewrite the target series using the change of index and the sum result from the previous part.

From the 'Hence' part, we know $\displaystyle \sum_{n=1}^N u_n = \sqrt{N^2+2N}$.
The required sum is $\displaystyle \sum_{n=2}^N u_{n-1}$.
Letting $k = n-1$, the limits change from $n=2 \to k=1$ and $n=N \to k=N-1$:
$\sum_{n=2}^N u_{n-1} = \sum_{k=1}^{N-1} u_k$
Using our known sum formula, substituting $N-1$ for the upper limit:
$= \sqrt{(N-1)^2+2(N-1)}$
$= \sqrt{N^2-2N+1+2N-2}$
$= \sqrt{N^2-1}$

1. Relate the required inequality to the exact series sum found in the earlier part.

From the earlier part, we established that:
$\displaystyle \sum_{n=1}^N \frac{2n+1}{\sqrt{n^2+2n}+\sqrt{n^2-1}} = \sqrt{N^2+2N}$

To prove that $\displaystyle \sum_{n=1}^N \frac{2n+1}{2n-1} > \sqrt{N^2+2N}$, we can compare the general terms of the two sums. It suffices to show that for all $n \ge 1$:
$\displaystyle \frac{2n+1}{2n-1} > \frac{2n+1}{\sqrt{n^2+2n}+\sqrt{n^2-1}}$

2. Compare the denominators by bounding the square root expressions.

Observe that for $n \ge 1$, we can bound the terms in the denominator of the right-hand expression:
Since $2n > 0$, we have $n^2+2n > n^2$, which strictly implies $\sqrt{n^2+2n} > n$.
Also, $n^2-1 \ge n^2-2n+1 = (n-1)^2$. Since $n \ge 1 \implies n-1 \ge 0$, we have $\sqrt{n^2-1} \ge n-1$.

Adding these two inequalities together gives:
$\sqrt{n^2+2n} + \sqrt{n^2-1} > n + (n-1) = 2n-1$

3. Deduce the inequality for the general terms and sum them to complete the proof.

Since $2n-1 > 0$ for all $n \ge 1$, taking the reciprocal reverses the inequality:
$\displaystyle \frac{1}{2n-1} > \frac{1}{\sqrt{n^2+2n}+\sqrt{n^2-1}}$

Multiplying both sides by $2n+1$ (which is strictly positive) gives:
$\displaystyle \frac{2n+1}{2n-1} > \frac{2n+1}{\sqrt{n^2+2n}+\sqrt{n^2-1}}$

Summing both sides from $n=1$ to $N$ yields:
$\displaystyle \sum_{n=1}^N \frac{2n+1}{2n-1} > \sum_{n=1}^N \frac{2n+1}{\sqrt{n^2+2n}+\sqrt{n^2-1}}$

Substituting the exact sum evaluated earlier into the right-hand side, we obtain:
$\displaystyle \sum_{n=1}^N \frac{2n+1}{2n-1} > \sqrt{N^2+2N} \quad \text{(Shown)}$