Find the sum of the series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^5} = a_n$ within three decimal places.
- The sum is estimated by $\displaystyle a_n \approx \sum_{k=1}^{n}\frac{1}{k^5}+R(n)$ where $R(n)$ is given by some value chosen between $\displaystyle \lim_{L \to \infty}\int_{n}^{L}\frac{1}{x^5}{\ \mathrm{d}x}$ and $\displaystyle \lim_{L \to \infty}\int_{n+1}^{L}\frac{1}{x^5}{\ \mathrm{d}x}$ inclusive, such that $\displaystyle \lim_{L \to \infty}\int_{n}^{L}\frac{1}{x^5}{\ \mathrm{d}x}$ is the maximum amount of error possible, and $\displaystyle \lim_{L \to \infty}\int_{n+1}^{L}\frac{1}{x^5}{\ \mathrm{d}x}$ is the minimum.
- Find: $\displaystyle \frac{1}{1000} > \lim_{L \to \infty}\left.\frac{4}{x^4}\right\vert_{L}^{x} \implies \frac{1}{4000} > \frac{1}{x^4}-\lim_{L \to \infty}\frac{1}{L^4} \implies x^4 > 4000 \implies x \approx 8 \implies n=8$
- $a_n \approx \displaystyle \sum_{n=1}^{8}\frac{1}{n^5} + \lim_{L \to \infty}\int_{8}^{L}\frac{1}{x^5}{\ \mathrm{d}x}$
Is this approach to estimating the sum of the series appropriate?