1. **Problem statement:** (a)(i) Given the sequence $x_n = \frac{4n^2 + 3n + 99.5}{2n^2 + 5}$, find the limit $\ell$ as $n \to \infty$ assuming the sequence converges. (a)(ii) Determine how large $n$ must be so that $|x_n - \ell| < 0.1$. (c) Find the values of $x \in \mathbb{R}$ for which the series $\sum_{k=2}^\infty \left(\frac{x+3}{2}\right)^k$ converges and find the sum of the series for those values. --- 2. **Step (a)(i): Find the limit $\ell$ of the sequence $x_n$** The sequence is given by: $$x_n = \frac{4n^2 + 3n + 99.5}{2n^2 + 5}$$ For large $n$, the highest degree terms dominate. The limit of a rational function where numerator and denominator are polynomials is the ratio of the leading coefficients if degrees are equal. Leading term numerator: $4n^2$ Leading term denominator: $2n^2$ So, $$\ell = \lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{4n^2 + 3n + 99.5}{2n^2 + 5} = \frac{4}{2} = 2$$ --- 3. **Step (a)(ii): Find $n$ such that $|x_n - \ell| < 0.1$** We want: $$|x_n - 2| < 0.1$$ Calculate $x_n - 2$: $$x_n - 2 = \frac{4n^2 + 3n + 99.5}{2n^2 + 5} - 2 = \frac{4n^2 + 3n + 99.5 - 2(2n^2 + 5)}{2n^2 + 5} = \frac{4n^2 + 3n + 99.5 - 4n^2 - 10}{2n^2 + 5} = \frac{3n + 89.5}{2n^2 + 5}$$ Since $n$ is natural and large, $x_n - 2 > 0$, so absolute value is just: $$\frac{3n + 89.5}{2n^2 + 5} < 0.1$$ Multiply both sides by denominator (positive): $$3n + 89.5 < 0.1(2n^2 + 5) = 0.2n^2 + 0.5$$ Rearranged: $$0.2n^2 - 3n - 89 < 0$$ Multiply by 5 to clear decimals: $$n^2 - 15n - 445 < 0$$ Solve quadratic inequality: Roots of $n^2 - 15n - 445 = 0$ are: $$n = \frac{15 \pm \sqrt{15^2 + 4 \times 445}}{2} = \frac{15 \pm \sqrt{225 + 1780}}{2} = \frac{15 \pm \sqrt{2005}}{2}$$ Approximate $\sqrt{2005} \approx 44.78$: $$n_1 = \frac{15 - 44.78}{2} \approx -14.89 \quad (\text{discard negative})$$ $$n_2 = \frac{15 + 44.78}{2} \approx 29.89$$ Since parabola opens upward, inequality $<0$ holds between roots, so for $n$ between $-14.89$ and $29.89$. Since $n$ is natural, $n < 30$ satisfies inequality. Check $n=30$: $$x_{30} - 2 = \frac{3(30) + 89.5}{2(30)^2 + 5} = \frac{90 + 89.5}{1805} = \frac{179.5}{1805} \approx 0.0995 < 0.1$$ So for $n \geq 30$, $|x_n - 2| < 0.1$. --- 4. **Step (c): Find values of $x$ for which the series converges and find the sum** The series is: $$\sum_{k=2}^\infty \left(\frac{x+3}{2}\right)^k$$ This is a geometric series with ratio: $$r = \frac{x+3}{2}$$ A geometric series converges if and only if $|r| < 1$: $$\left|\frac{x+3}{2}\right| < 1 \implies |x+3| < 2$$ This gives: $$-2 < x + 3 < 2 \implies -5 < x < -1$$ Sum of geometric series starting at $k=2$ is: $$S = \sum_{k=2}^\infty r^k = r^2 + r^3 + r^4 + \cdots = \frac{r^2}{1 - r}$$ So, $$S = \frac{\left(\frac{x+3}{2}\right)^2}{1 - \frac{x+3}{2}} = \frac{\frac{(x+3)^2}{4}}{\frac{2 - (x+3)}{2}} = \frac{(x+3)^2}{4} \times \frac{2}{2 - x - 3} = \frac{(x+3)^2}{4} \times \frac{2}{-x -1} = \frac{(x+3)^2}{2(-x -1)} = \frac{(x+3)^2}{-2(x+1)}$$ Simplify denominator sign: $$S = -\frac{(x+3)^2}{2(x+1)}$$ --- **Final answers:** - (a)(i) $\ell = 2$ - (a)(ii) For $n \geq 30$, $|x_n - 2| < 0.1$ - (c) The series converges for $x \in (-5, -1)$ and the sum is: $$S = -\frac{(x+3)^2}{2(x+1)}$$