MATH101 April 2012
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Question 04 (c)

Determine, with explanation, whether the series
 $\sum _{n=2}^{\infty }{\frac {(1)^{n}}{n(\ln n)^{101}}}$
converges absolutely, converges conditionally, or diverges.

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

Can you integrate ${\frac {1}{x(\ln x)^{101}}}$ ?

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Solution

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We will use the integral test. Note that
 $\int _{2}^{\infty }{\frac {dx}{x\ln(x)^{101}}}=\lim _{a\to \infty }\int _{2}^{a}{\frac {dx}{x\ln(x)^{101}}}$
We can calculate $\int _{2}^{a}{\frac {dx}{x\ln(x)^{101}}}$ by substitution with $u=\ln x$ (then du=(1/x)dx) to get
 ${\begin{aligned}\int _{2}^{a}{\frac {dx}{x\ln(x)^{101}}}&=\int _{\ln(2)}^{\ln(a)}u^{101}du\\&=\left.{\frac {1}{100u^{100}}}\right_{\ln(2)}^{\ln(a)}\\&={\frac {1}{100}}\left({\frac {1}{(\ln(a))^{100}}}{\frac {1}{(\ln(2))^{100}}}\right)\end{aligned}}$
Hence
 $\int _{2}^{\infty }{\frac {dx}{x\ln(x)^{101}}}=\lim _{a\to \infty }{\frac {1}{100}}\left({\frac {1}{\ln(2)^{100}}}{\frac {1}{\ln(a)^{100}}}\right)={\frac {1}{100\ln(2)^{100}}}$.
This implies that the series converges absolutely by the integral test.

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