A geometric series has first term $2\sqrt{2}\sin\theta$ and second term $\sin 2\theta$. Show that the series is convergent for $\tfrac{\pi}{2} < \theta < \pi$. It is given that $\theta = \tfrac{3\pi}{4}$ and $S_n$ denotes the sum of the first $n$ terms of the series. Find $S_n$ and hence determine the exact value of $S_{\infty}$. Show that $\displaystyle \frac{1}{\sqrt{n+1}+\sqrt{n}} = \sqrt{n+1} - \sqrt{n}$, where $n \in \mathbb{Z}^{+}$. Hence find the least possible value of $N$ such that $\displaystyle \sum_{n=4}^{N} \frac{1}{\sqrt{n+1}+\sqrt{n}} > 100$.

1. Find the common ratio $r$ in its simplest form.

$r = \frac{\text{second term}}{\text{first term}} = \frac{\sin 2\theta}{2\sqrt{2}\sin\theta}$
Using the double angle identity $\sin 2\theta = 2\sin\theta\cos\theta$:
$r = \frac{2\sin\theta\cos\theta}{2\sqrt{2}\sin\theta} = \frac{\cos\theta}{\sqrt{2}}$

2. Use the given domain of $\theta$ to show that $|r| < 1$.

For the given interval $\frac{\pi}{2} < \theta < \pi$, the cosine function is bounded by $-1 < \cos\theta < 0$.
Therefore, dividing by $\sqrt{2}$ gives:
$-\frac{1}{\sqrt{2}} < \frac{\cos\theta}{\sqrt{2}} < 0$
Since $-\frac{1}{\sqrt{2}} \approx -0.707$, we have $-1 < r < 0$.
Because $|r| < 1$, the geometric series is convergent.

1. Evaluate the first term $a$ and common ratio $r$ when $\theta = \frac{3\pi}{4}$.

The first term is $a = 2\sqrt{2}\sin\left(\frac{3\pi}{4}\right)$.
Since $\sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}$,
$a = 2\sqrt{2}\left(\frac{\sqrt{2}}{2}\right) = 2$

The second term is $\sin\left(2\left(\frac{3\pi}{4}\right)\right) = \sin\left(\frac{3\pi}{2}\right) = -1$.
So, the common ratio $r = \frac{-1}{2} = -\frac{1}{2}$.

2. Substitute $a$ and $r$ into the formula for the sum of the first $n$ terms, $S_n$.

$S_n = \frac{a(1 - r^n)}{1 - r}$
$S_n = \frac{2\left(1 - \left(-\frac{1}{2}\right)^n\right)}{1 - \left(-\frac{1}{2}\right)}$
$S_n = \frac{2\left(1 - \left(-\frac{1}{2}\right)^n\right)}{\frac{3}{2}}$
$S_n = \frac{4}{3}\left(1 - \left(-\frac{1}{2}\right)^n\right)$

3. Determine the sum to infinity $S_\infty$ using the result for $S_n$.

As $n \to \infty$, since $|-\frac{1}{2}| < 1$, the term $\left(-\frac{1}{2}\right)^n \to 0$.
Hence,
$S_\infty = \lim_{n\to\infty} S_n = \frac{4}{3}(1 - 0) = \frac{4}{3}$

1. Multiply the numerator and denominator by the conjugate of the denominator.

$\frac{1}{\sqrt{n+1}+\sqrt{n}} = \frac{1}{\sqrt{n+1}+\sqrt{n}} \times \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}$

2. Simplify the denominator using the difference of squares identity to prove the result.

$= \frac{\sqrt{n+1}-\sqrt{n}}{(\sqrt{n+1})^2 - (\sqrt{n})^2}$\n$= \frac{\sqrt{n+1}-\sqrt{n}}{(n+1) - n}$\n$= \sqrt{n+1}-\sqrt{n} \quad \text{(shown)}$

1. Use the identity from part (b)(i) to rewrite the terms of the series.

$\sum_{n=4}^{N} \frac{1}{\sqrt{n+1}+\sqrt{n}} = \sum_{n=4}^{N} (\sqrt{n+1} - \sqrt{n})$

2. Evaluate the sum using the method of differences.

3. Set up and solve the inequality to find the least integer $N$.

Since $N$ must be an integer, the least possible value is $10404$.