Question

I completely confused. Estimate the value of series 1 to infinity 6/n^3 to within .048 of its exact value

Answer 1

\[\int_{1}^{\infty} \frac{6}{x^3}dx \]

Answer 2

\[6\int\limits_{1}^{t}x^{-3}dx=-3[x^{-2}] = -3\lim_{t \rightarrow 0.48}[\frac{1}{0.48}-1] = \]

Answer 3

why did you have .48 plugged in for t