Problem source

\begin{questionparts}
\itemSuppose 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 $n\leqslant4$. 
\item 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 ${\displaystyle {\displaystyle \sum_{n=1}^{N}\frac{1}{n}>10.}}$
\end{questionparts}

Solution source

\begin{questionparts}
\item 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.
\item 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 $1$.
\end{questionparts}