Suppose that the real number \(x\) satisfies the \(n\) inequalities
\begin{alignat*}{2}
1<\ & x & & < 2\\
2<\ & x^{2} & & < 3\\
3<\ & x^{3} & & < 4\\
& \vdots\\
n<\ & x^{n} & & < n+1
\end{alignat*}
Prove without the use of a calculator that \(n\leqslant4\).
If \(n\) is an integer strictly greater than 1, by considering
how many terms there are in
\[
\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n^{2}},
\]
or otherwise, show that
\[
\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{n^{2}}>1.
\]
Hence or otherwise find, with justification, an integer \(N\) such
that \({\displaystyle {\displaystyle \sum_{n=1}^{N}\frac{1}{n}>10.}}\)
Solution:
Suppose \(n > 4\) then the following inequalities are both true
\begin{align*}
3 < x^3 < 4 & \Rightarrow 3^5 < x^{15} < 4^{5}\\
5 < x^5 < 6 & \Rightarrow 5^{3} < x^{15} < 6^3
\end{align*}
But \(3^5 = 243\) and \(6^3 = 216\) so \(243 < x^{15} < 216\) whichis a contradiction.
This question is wrong. Consider \(n = 2\), then \(\frac{1}{2+1} + \frac{1}{2+2} = \frac13+\frac14 = \frac{7}{12} < 1\). The question should be about \(n \geq 4\).
\begin{align*}
\frac{1}{n+1}+\frac1{n+2}+\cdots + \frac{1}{2n} > \frac{n}{2n} &= \frac12 \\
\frac{1}{2n+1}+\frac1{2n+2}+\cdots + \frac{1}{3n} > \frac{n}{3n} &= \frac13 \\
\frac{1}{4n+1}+\frac1{4n+2}+\cdots + \frac{1}{4n} > \frac{n}{4n} &= \frac14 \\
\sum_{k=1}^{n^2-n} \frac{1}{n+k} > \frac{13}{12} &> 1
\end{align*}
We have a stronger result, \(\frac1{n+1} + \cdots + \frac1{4n} > 1\) for \(n > 4\)
so we can take \(N = 4^{10}\) since, since there will be \(9\) sequences from \(\frac{1}{4^{i}+1} \to \frac{1}{4^{i+1}}\) and we will have \(\frac1{1}\) at the start to give use the extra \(1\).