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

MATH101 April 2013

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[hide] 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.

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[show] Hint 1
What hypotheses are required for the Integral Test? Which one fails?

[show] 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.

[show] 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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[show] 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 x_n of values that grow with no limit, and for which ƒ'(x_n) 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'(x_n) positive is to choose x_n such that the positive term is maximal. This is achieved when cos(x_n) = 1. In this case, sin(x_n) = 0 and we find that the enumerator becomes 2(1+x_n²)-2x_n² = 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.