Estimating the Value of a Series — Question 3

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Question 3

Let S=∑n=1∞1/n3\displaystyle S=\sum_{n=1}^{\infty}1/n^3. Using the Integral Test upper remainder bound, find the least final index NN certified to give absolute error |RN|<1210−4|R_N|<\tfrac12 10^{-4}. Explain whether this tolerance alone guarantees identical rounding to four decimal places.

Original worksheet page 1: question and worked solution for 4-13-003
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Question 3 – Solution

Step 1: Obtain an upper tail bound.

With f(x)=x−3f(x)=x^{-3}, 0<RN≤∫N∞x−3dx=[−12x2]N∞=12N2.0<R_N\le\int_N^{\infty}x^{-3}\,dx=\left[-\frac1{2x^2}\right]_N^{\infty}=\frac1{2N^2}.

Step 2: Impose the absolute-error tolerance.

The requested error is below 0.00005=5×10−50.00005=5\times10^{-5}: 12N2<5×10−5⇔N2>10,000⇔N>100.\frac1{2N^2}<5\times10^{-5}\iff N^2>10{,}000\iff N>100. Thus the least integer certified by this bound is N=101\boxed{N=101}.

Step 3: Check the strict endpoint.

At N=100N=100, the bound equals 5×10−55\times10^{-5}, so it does not guarantee a strict error smaller than half a unit in the fourth decimal place. At N=101N=101, it does.

This absolute-error tolerance alone does not guarantee identical rounding: two nearby numbers can lie on opposite sides of a rounding threshold.

Original worksheet page 2: question and worked solution for 4-13-003

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