The series converges

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Calculus of change and accumulation · Calculus of change and accumulation

Mathematical Methods for Physics, Ch Accumulation: integration and the techniques that make it possible

Problem

The series S=∑n=1∞1n3/2\displaystyle S = \sum_{n=1}^{\infty}\frac{1}{n^{3/2}} converges, since 3/2>13/2 > 1. ((a)) Use the staircase bracket of the integral test to show that 2≤S≤32 \le S \le 3. ((b)) That bracket is wide because the first few terms are large. The first ten terms add up to S10=∑n=110n−3/2=1.9953S_{10} = \sum_{n=1}^{10} n^{-3/2} = 1.9953. Apply the same staircase argument to the remaining terms ∑n=11∞n−3/2\sum_{n=11}^{\infty} n^{-3/2} alone, and so bracket SS in an interval narrower than 0.030.03. Take the midpoint of that interval as an estimate of SS, and state the largest error it can have.

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