2003 Paper 2 Q7

Year: 2003
Paper: 2
Question Number: 7

Course: UFM Pure
Section: Sequences and series, recurrence and convergence

Difficulty: 1600.0 Banger: 1500.0

integration integration by parts sequences series inequality summation logarithmic improper integral geometric series convergence

Problem

Show that, if $n>0\,$, then $$ \int_{e^{1/n}}^\infty\,{{\ln x} \over {x^{n+1}}}\,\d x = {2 \over {n^2\e}}\;. $$ You may assume that $\ds \frac{\ln x} x \to 0\;$ as $x\to\infty\,$. Explain why, if $1 < a < b\,$, then $$ \int_b^\infty\,{{\ln x} \over {x^{n+1}}}\,\d x < \int_a^\infty\,{{\ln x} \over {x^{n+1}}}\,\d x\;. $$ Deduce that $$ \sum_{n=1}^{N}{1 \over n^2} < {\e \over 2}\int_{\e^{1/N}}^{\infty} \left({1-x^{-N}} \over {x^2-x}\right) \ln x\,\d x\;, $$ where $N\,$ is any integer greater than $1$.

No solution available for this problem.

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Difficulty Rating: 1600.0

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Banger Rating: 1500.0

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Problem source

Show that, 
 if $n>0\,$, then
$$
\int_{e^{1/n}}^\infty\,{{\ln x} \over {x^{n+1}}}\,\d x
=  {2 \over  {n^2\e}}\;.  
$$
You may assume that $\ds \frac{\ln x} x \to 0\;$ as $x\to\infty\,$.

Explain why, if $1 < a < b\,$, then
$$
\int_b^\infty\,{{\ln x} \over {x^{n+1}}}\,\d x  
<
\int_a^\infty\,{{\ln x} \over {x^{n+1}}}\,\d x\;.
$$

Deduce that
$$
\sum_{n=1}^{N}{1 \over n^2} <
{\e \over 2}\int_{\e^{1/N}}^{\infty}
\left({1-x^{-N}} \over {x^2-x}\right) \ln x\,\d x\;,
$$
where  $N\,$ is any integer greater than $1$.

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