Science:Math Exam Resources/Courses/MATH101/April 2017/Question 02 (c) (i)

MATH101 April 2017

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c) (i)  •  Q2 (c) (ii)  •  Q2 (c) (iii)  •  Q2 (c) (iv)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  Q6 (a)  •  Q6 (b)  •  Q7 (a)  •  Q7 (b)  •  Q8  •  Q9  •  Q10  •  Q11 (a)  •  Q11 (b)  •  Q11 (c)  •  

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Question 02 (c) (i)

For each of the following series, choose the appropriate statement. (Write N, O, P, S, or T in each box; each answer will be used at most once, and each series matches a single answer only.) N: The series converges by the Ratio Test. O: The series converges absolutely by the Comparison Test with a p-series. P: The series converges by the Alternating Series Test. S: The series diverges. T: The series converges by the Integral Test. (i) $\sum _{n=2}^{\infty }{\frac {1}{n(\log n)^{2}}}$

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Look for a nice ${\textstyle u}$-substitution.

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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. Since the function ${\frac {1}{x(\log(x))^{2}}}$ is positive and decreasing on $[2,\infty )$, we can consider the integral test, According to the test, the series converges if $$\int _{2}^{\infty }{\frac {1}{x(\log(x))^{2}}}dx<\infty .$$ To evaluate this integral, substitute ${\textstyle u=\log(x)}$, ${\textstyle du={\frac {1}{x}}\,dx}$. This simplifies to $$\int _{\log(2)}^{\infty }{\frac {1}{u^{2}}}\,du=-u^{-1}{\Big |}_{\log(2)}^{\infty }=\lim _{u\rightarrow \infty }-u^{-1}+{\frac {1}{\log(2)}}={\frac {1}{\log(2)}}$$ This is indeed convergent, hence the answer is ${\textstyle \color {blue}{\text{T}}}$.