1999 Paper 1 Q8

Year: 1999
Paper: 1
Question Number: 8

Course: LFM Pure
Section: Integration

Difficulty: 1500.0 Banger: 1500.0

Problem

The function \(\f\) satisfies \(0\leqslant\f(t)\leqslant K\) when \(0\leqslant t\leqslant x\). Explain by means of a sketch, or otherwise, why \[0\leqslant\int_{0}^{x} \f (t)\,{\mathrm d}t \leqslant Kx.\] By considering \(\displaystyle \int_{0}^{1}\frac{t}{n(n-t)}\,{\mathrm d}t\), or otherwise, show that, if \(n>1\), \[ 0\le \ln \left( \frac n{n-1}\right) -\frac 1n \le \frac 1 {n-1} - \frac 1n \] and deduce that \[ 0\le \ln N -\sum_{n=2}^N \frac1n \le 1. \] Deduce that as \(N\to \infty\) \[ \sum_{n=1}^N \frac1n \to\infty. \] Noting that \(2^{10}=1024\), show also that if \(N<10^{30}\) then \[ \sum_{n=1}^N \frac1n <101. \]

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Problem source
The function $\f$ satisfies  $0\leqslant\f(t)\leqslant K$
when $0\leqslant t\leqslant x$. Explain by means of a sketch, or
otherwise,  why  
\[0\leqslant\int_{0}^{x} \f (t)\,{\mathrm d}t
\leqslant Kx.\]
By considering 
$\displaystyle \int_{0}^{1}\frac{t}{n(n-t)}\,{\mathrm d}t$, or otherwise,
show that, if $n>1$,
\[
0\le \ln \left( \frac n{n-1}\right)  -\frac 1n \le \frac 1 {n-1} - \frac 1n
\]
and deduce that
\[
0\le \ln N -\sum_{n=2}^N \frac1n  \le 1.
\]
Deduce that as $N\to \infty$
\[
 \sum_{n=1}^N \frac1n \to\infty.
\]
Noting that $2^{10}=1024$, show also that  if $N<10^{30}$ then
\[
 \sum_{n=1}^N \frac1n <101.
\]