Science:Math Exam Resources/Courses/MATH101/April 2011/Question 01 (g)

MATH101 April 2011

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q1 (h)  •  Q1 (i)  •  Q1 (j)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q4 (d)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q6 (a)  •  Q6 (b)  •  Q7  •  Q8  •  

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Question 01 (g)

For this question, the exam asks to put your answer in the box provided but also to show your work. This question is worth 3 marks. Full marks is given for correct answers placed in the box and at most 1 mark is given for an incorrect answer. You are required to simplify your answers as much as possible. How many terms of the convergent series $\sum _{n=1}^{\infty }n^{-3}$ are needed to ensure that the sum is accurate to within 1/20,000?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
This series can be thought of as a Riemann sum for a certain integral.

Hint 2
We want to find the smallest k such that $\left|\sum _{n=1}^{\infty }{\frac {1}{n^{3}}}-\sum _{n=1}^{k}{\frac {1}{n^{3}}}\right|\leq {\frac {1}{20000}}.$ Or, equivalently, $\left|\sum _{n=k+1}^{\infty }{\frac {1}{n^{3}}}\right|\leq {\frac {1}{20000}}.$ Interpret this sum as a Riemann sum of a certain integral that you can compute.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We want to find the smallest k such that $\left|\sum _{n=1}^{\infty }{\frac {1}{n^{3}}}-\sum _{n=1}^{k}{\frac {1}{n^{3}}}\right|\leq {\frac {1}{20000}}.$ Or, equivalently, $\left|\sum _{n=k+1}^{\infty }{\frac {1}{n^{3}}}\right|\leq {\frac {1}{20000}}.$ Since all of the terms in the sum are positive, the sum itself is positive. So, the absolute value signs are unnecessary. Note that $\sum _{n=k+1}^{\infty }{\frac {1}{n^{3}}}$ can be interpreted as the right Riemann sum for the integral $\int _{k}^{\infty }{\frac {1}{x^{3}}}dx$. Since $1/x^{3}$ is monotonically decreasing, the right Riemann sum is an underestimation for the area under the curve. As step size for the Riemann sum $\Delta x=1$ is used. Therefore, we have $\sum _{n=k+1}^{\infty }{\frac {1}{n^{3}}}<\int _{k}^{\infty }{\frac {1}{x^{3}}}dx,$ and hence it suffices to find the minimal value of $k$ such that the integral above is ≤ 20,000. But the integral $\int _{k}^{\infty }{\frac {1}{x^{3}}}dx=\left[-{\frac {1}{2x^{2}}}\right]_{k}^{\infty }={\frac {1}{2k^{2}}}.$ Hence $\sum _{n=k+1}^{\infty }{\frac {1}{n^{3}}}<\int _{k}^{\infty }{\frac {1}{x^{3}}}dx={\frac {1}{2k^{2}}}\leq {\frac {1}{20000}}.$ The minimal such $k$ is $k=100$ (in fact, for this choice of $k$, the right ≤ becomes an $=$). Therefore we must take the first 100 terms of the sum for the approximation to be accurate to within 1/20000.

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