Integral Test

Calculus Level 3

Using the integral test , find the best possible estimation for the series $\displaystyle\sum_{n=1}^\infty \frac{1}{n^3}$ using a partial sum with $10$ terms.

Note that $\displaystyle\sum_{n=1}^{10} \frac{1}{n^3} \approx 1.19753$ and $\tfrac{1}{11^2} \approx 0.00826$ .

Value 1.19753, error at most 0.01 Value 1.20666, error at most 0.00087 Value 1.2021, error at most 0.00044 Value 1.20753, error at most 0.00413 Value 1.19753, error at most 0.00826

1 solution

Henry Maltby
May 18, 2016

Note that $1.19753 + \int_{11}^\infty \frac{1}{x^3} \, dx < \sum_{n=1}^\infty \frac{1}{n^3} < 1.19753 + \int_{10}^\infty \frac{1}{x^3} \, dx.$ Because $\displaystyle\int \frac{1}{x^3} \, dx = - \frac{1}{2x^2} + c$ , the interval becomes $1.19753 + 0.00413 < \sum_{n=1}^\infty \frac{1}{n^3} < 1.19753 + 0.005.$ It follows that the sum in question is approximately $1.19753 + 0.5 \cdot (0.00413 + 0.005) = 1.202095$ with an error of at most $0.5 \cdot (0.005 - 0.00413) = 0.000435$ . Therefore, the answer is that the series has value $1.2021$ with an error of at most $0.00044$ .

The actual value of the series is $\sum_{n=1}^\infty = 1.202056903\dots .$

Wow, that was a great explanation!

Hayden Park - 11 months, 1 week ago

Integral Test

1 solution

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