1. **Problem statement:** Show that the percentage error in approximating the sum to infinity of a geometric series by the sum of the first $n$ terms is given by $$\text{Percentage error} = \frac{100 x^{n+1}}{1 - x^{n+1}}$$ where $|x| < 1$. 2. **Formula for sum to infinity:** For a geometric series with first term 1 and common ratio $x$, the sum to infinity is $$S_\infty = \frac{1}{1 - x}$$ provided $|x| < 1$. 3. **Sum of first $n$ terms:** The sum of the first $n$ terms is $$S_n = \frac{1 - x^{n}}{1 - x}$$ 4. **Error in approximation:** The error when approximating $S_\infty$ by $S_n$ is $$E = S_\infty - S_n = \frac{1}{1 - x} - \frac{1 - x^{n}}{1 - x} = \frac{x^{n}}{1 - x}$$ 5. **Percentage error:** The percentage error relative to the sum to infinity is $$\text{Percentage error} = 100 \times \frac{E}{S_\infty} = 100 \times \frac{\frac{x^{n}}{1 - x}}{\frac{1}{1 - x}} = 100 x^{n}$$ 6. **Correction to the problem statement:** The problem states the error as $$\frac{100 x^{n+1}}{1 - x^{n+1}}$$ which suggests the error is relative to the partial sum including $n+1$ terms. Using the sum to $n+1$ terms, $$S_{n+1} = \frac{1 - x^{n+1}}{1 - x}$$ and the error is $$E = S_\infty - S_{n+1} = \frac{x^{n+1}}{1 - x}$$ Percentage error relative to $S_{n+1}$ is $$100 \times \frac{E}{S_{n+1}} = 100 \times \frac{\frac{x^{n+1}}{1 - x}}{\frac{1 - x^{n+1}}{1 - x}} = \frac{100 x^{n+1}}{1 - x^{n+1}}$$ 7. **Deduction for $n$ given error less than $k\%$:** We want $$\frac{100 x^{n+1}}{1 - x^{n+1}} < k$$ Rearranging, $$100 x^{n+1} < k (1 - x^{n+1})$$ $$100 x^{n+1} + k x^{n+1} < k$$ $$x^{n+1} (100 + k) < k$$ Taking logarithms, $$ (n+1) \log x < \log \frac{k}{100 + k}$$ Since $0 < x < 1$, $\log x < 0$, dividing by $\log x$ reverses inequality: $$ n + 1 > \frac{\log \frac{k}{100 + k}}{\log x}$$ or equivalently, $$ n > \frac{\log (k/(100 + k))}{\log x} - 1$$ 8. **Evaluate for $k=1$ and $x=0.9$:** Calculate $$ n > \frac{\log \frac{1}{101}}{\log 0.9} - 1$$ Using natural logs, $$\log \frac{1}{101} = -\log 101 \approx -4.615$$ $$\log 0.9 \approx -0.10536$$ So, $$ n > \frac{-4.615}{-0.10536} - 1 = 43.82 - 1 = 42.82$$ Thus, $$ n \geq 43$$ **Final answers:** - Percentage error formula: $$\frac{100 x^{n+1}}{1 - x^{n+1}}$$ - Condition on $n$: $$ n > \frac{\log (k/(100 + k))}{\log x} - 1$$ - For $k=1$, $x=0.9$: $$ n \geq 43$$