Science:Math Exam Resources/Courses/MATH105/April 2012/Question 08 (a)

MATH105 April 2012

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Question 08 (a)

Short-answer questions. No credit will be given for the answer (even if it is correct) without the accompanying work. The Maclaurin series for arctan x is given by $\arctan x=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{2n+1}}{2n+1}}$ which has radius of convergence equal to 1. Use this fact to compute the exact value of the series below: $\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)3^{n}}}$ (Hint: Note that $3^{n}=({\sqrt {3}})^{2n}$).

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
By a visual inspection, we see that the terms that don't match up to the expression for $\arctan(x)$ are the terms corresponding to $x^{2n+1}$ and to $3^{-n}=(3^{-1})^{n}$ (moving the denominator up to the numerator). Thus, we want to get these two terms to match. How can we rearrange these terms to achieve this goal?

Hint 2
Utilize the hint given and note $3^{-n}=(3^{-1})^{n}=({\sqrt {3}}^{-1})^{2n}={\sqrt {3}}{\sqrt {3}}^{-1}({\sqrt {3}}^{-1})^{2n}={\sqrt {3}}({\sqrt {3}}^{-1})^{2n+1}$

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. To evaluate $S=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)3^{n}}}$, we need to make it look more like $\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{2n+1}}{2n+1}}$ for some x. Both sums start with $n=0$, have $(-1)^{n}$, and have a $2n+1$ in the denominator. Therefore, we need that $x^{2n+1}={\frac {1}{3^{n}}}$. Following the hint, we write ${\frac {1}{3^{n}}}={\frac {1}{({\sqrt {3}})^{2n}}}$ which we can make even more similar to $x^{2n+1}$ by writing it as $({\frac {1}{\sqrt {3}}})^{2n}$. Let's write out the two sums again and see how close we are: $\arctan x=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{2n+1}}{2n+1}}$ and we wish to evaluate $S=\sum _{n=0}^{\infty }(-1)^{n}{\frac {(1/{\sqrt {3}})^{2n}}{2n+1}}$. If $x=1/{\sqrt {3}}$, the only difference between the two sums now is a factor of $(1/{\sqrt {3}})^{1}$, which doesn't depend on n. We can multiply S by $1/{\sqrt {3}}$ (and put that factor in the sum) and also divide S by that same factor (keeping it outside the sum): ${\begin{aligned}S&={\frac {1}{1/{\sqrt {3}}}}\sum _{n=0}^{\infty }(-1)^{n}{\frac {(1/{\sqrt {3}})^{2n+1}}{2n+1}}\\&{\stackrel {{\text{recognize sum as}}\arctan(1/{\sqrt {3}})}{=}}{\sqrt {3}}\arctan(1/{\sqrt {3}})\\&={\sqrt {3}}{\frac {\pi }{6}}\end{aligned}}$

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