Mathematics

Mathematics

Problem

The number e−1e^{-1} can be written as the alternating series e−1=∑n=0∞(−1)n 1n!=1−1+12−16+124−1120+… .e^{-1}=\sum_{n=0}^{\infty}(-1)^{n}\,\frac{1}{n!}=1-1+\frac12-\frac16+\frac1{24}-\frac1{120}+\dots. Let P4P_4 denote the fourth–degree truncation (through the n=4n=4 term) of the series. a. Write P4P_4 explicitly and evaluate it to six decimal places. b. Compute the magnitude of the next omitted term, ∣a5∣|a_5|, and give its decimal value. c. Determine whether P4P_4 approximates e−1e^{-1} to within 0.00010.0001. Justify your answer numerically. d. Find the least positive integer NN such that the magnitude of the (N+1)(N+1)st term is less than 0.000050.00005. State the value of NN and explain your reasoning.

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