Science:Math Exam Resources/Courses/MATH101/April 2013/Question 12 (b)

MATH101 April 2013

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Question 12 (b)
Full-Solution Problem. Justify your answer and show all your work. Simplification of numerical answers is required.. Explain why you cannot conclude that $\displaystyle \sum _{n=2}^{\infty }{\frac {n+\sin n}{1+n^{2}}}$ diverges from part (a) and the Integral Test.

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
What hypotheses are required for the Integral Test? Which one fails?

Hint 2
If $\displaystyle f(x)={\frac {x+\sin x}{1+x^{2}}}$, then its clear it is positive, tends to 0, and agrees with the summands on the natural numbers. So prove it is never eventually decreasing.

Hint 3
It might be easiest to prove $\displaystyle f(x)={\frac {x+\sin x}{1+x^{2}}}$ is never eventually decreasing by differentiating and using a differential calculus result.

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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. If $\displaystyle f(x)={\frac {x+\sin x}{1+x^{2}}}$, then its clear it is positive, tends to 0, and agrees with the summands on the natural numbers. So we should prove it is never eventually decreasing. Taking the derivative, $\displaystyle f'(x)={\frac {(1+\cos x)(1+x^{2})-2x(x+\sin x)}{(1+x^{2})^{2}}}$ We want to show that ƒ is never eventually decreasing. For this, it suffices to find a sequence xn of values that grow with no limit, and for which ƒ'(xn) is positive. How would be go about finding such a sequence? Let us look at the enumerator more carefully. Our best bet to make ƒ'(xn) positive is to choose xn such that the positive term is maximal. This is achieved when cos(xn) = 1. In this case, sin(xn) = 0 and we find that the enumerator becomes 2(1+xn2)-2xn2 = 2 > 0. Hence, we choose $\displaystyle x_{n}=2n\pi$ for any integer n, and double check that indeed the numerator becomes: $\displaystyle {\begin{aligned}&(1+\cos(2n\pi ))(1+(2n\pi )^{2})-2(2n\pi )((2n\pi )+\sin(2n\pi ))\\=\ &2(1+4n^{2}\pi ^{2})-8n^{2}\pi ^{2}\\=\ &2>0\end{aligned}}$ The denominator is always positive, so $\displaystyle f'(x_{n})>0$ for these values of $\displaystyle x_{n}$, so that $\displaystyle f(x_{n})$ is increasing at these values of $\displaystyle x_{n}$. Since $\displaystyle 2n\pi$ grows with no limit, $\displaystyle f(x)$ is never eventually decreasing, and hence the integral test fails.

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