Year: 2012
Paper: 2
Question Number: 4
Course: UFM Pure
Section: Taylor series
There were just over 1000 entries for paper II this year, almost exactly the same number as last year. Overall, the paper was found marginally easier than its predecessor, which means that it was pitched at exactly the level intended and produced the hoped-for outcomes. Almost 50 candidates scored 100 marks or more, with more than 400 gaining at least half marks on the paper. At the lower end of the scale, around a quarter of the entry failed to score more than 40 marks. It was pleasing to note that the advice of recent years, encouraging students not to make attempts at lots of early parts to questions but rather to spend their time getting to grips with the six that can count towards their paper total, was more obviously being heeded in 2012 than I can recall being the case previously. As in previous years, the pure maths questions provided the bulk of candidates' work, with relatively few efforts to be found at the applied ones. Questions 1 and 2 were the most popular questions, although each drew only around 800 "hits" – fewer than usual. Questions 3 – 5 & 8 were almost as popular (around 700), with Q6 attracting the interest of under 450 candidates and Q7 under 200. Q9 was the most popular applied question – and, as it turned out, the most successfully attempted question on the paper – with very little interest shown in the rest of Sections B or C.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
In this question, you may assume that the infinite series
\[
\ln(1+x) = x-\frac{x^2}2 + \frac{x^3}{3} -\frac {x^4}4 +\cdots
+ (-1)^{n+1} \frac {x^n}{n} + \cdots
\]
is valid for $\vert x \vert <1$.
\begin{questionparts}
\item
Let $n$ be an integer greater than 1. Show that, for any positive integer $k$,
\[
\frac1{(k+1)n^{k+1}}
<
\frac1{kn^{k}}\,.
\]
Hence show that $\displaystyle \ln\! \left(1+\frac1n\right) <\frac1n\,$. Deduce that
\[
\left(1+\frac1n\right)^{\!n}<\e\,.
\]
\item
Show, using an expansion in powers of $\dfrac1y\,$, that $ \displaystyle
\ln \! \left(\frac{2y+1}{2y-1}\right)
> \frac 1y
%= \sum _{r=0}^\infty \frac 1{(2r+1)(2y)^{2r}}\,.
$ for $y>\frac12$.
Deduce that, for any positive integer $n$,
\[
\e < \left(1+\frac1n\right)^{\! n+\frac12}\,.
\]
\item Use parts (i) and (ii) to show that as $n\to\infty$
\[
\left(1+\frac1n\right)^{\!n}
\to \e\,.
\]
\end{questionparts}\begin{questionparts}
\item Since $k \geq 1$ we have $n^{k+1} > n^k$ and $(k+1) > k$, therefore $(k+1)n^{k+1} >kn^k \Rightarrow \frac{1}{(k+1)n^{k+1}} < \frac{1}{kn^k}$
\begin{align*}
&& \ln \left ( 1 + \frac1n \right) &= \frac1n -\frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \cdots \\
&&&= \frac1n - \underbrace{\left (\frac{1}{2n^2}-\frac{1}{3n^3} \right)}_{>0}- \underbrace{\left (\frac{1}{4n^4}-\frac{1}{5n^5} \right)}_{>0} - \cdot \\
&&&< \frac1n \\
\\
\Rightarrow && n \ln \left ( 1 + \frac1n \right) &< 1 \\
\Rightarrow && \ln \left ( \left ( 1 + \frac1n \right)^n \right) &< 1 \\
\Rightarrow && \left ( 1 + \frac1n \right)^n &< e
\end{align*}
\item $\,$
\begin{align*}
&&\ln \left(\frac{2y+1}{2y-1}\right) &= \ln \left (1 + \frac{1}{2y} \right)-\ln \left (1 - \frac{1}{2y} \right) \\
&&&= \frac{1}{2y} - \frac{1}{2(2y)^2} + \frac{1}{3(2y)^3} - \cdots - \left (-\frac{1}{2y} - \frac{1}{2(2y)^2} - \frac{1}{3(2y)^3} - \cdots \right) \\
&&&= \frac{1}{y} + \frac{2}{3(2y)^3} + \frac{2}{5(2y)^5} \\
&&&= \sum_{r=1}^{\infty} \frac{2}{(2r-1)(2y)^{2r-1}} \\
&&&> \frac1y \\
\\
\Rightarrow && \ln \left (1 + \frac{1}{y-\frac12} \right) &> \frac{1}{y} \\\Rightarrow && \ln \left (1 + \frac{1}{n} \right) &> \frac{1}{n+\frac12} \\
\Rightarrow &&(n+\tfrac12) \ln \left (1 + \frac{1}{n} \right) &> 1\\
\Rightarrow && \ln \left ( \left (1 + \frac{1}{n} \right)^{n+\tfrac12} \right) &> 1\\
\Rightarrow && \left (1 + \frac{1}{n} \right)^{n+\tfrac12} & > e
\end{align*}
\end{questionparts}
\item Since $\left (1 + \frac1n \right)^n$ is both bounded above, and increasing, it must tend to some limit $L$.
\begin{align*}
&& \lim_{n \to \infty} \left (1 + \frac1n \right)^n && \leq e &\leq & \lim_{n \to \infty} \left (1 + \frac1n \right)^{n+\frac12} \\
\Rightarrow && \lim_{n \to \infty} \left (1 + \frac1n \right)^n && \leq e &\leq & \lim_{n \to \infty} \left (1 + \frac1n \right)^{n} \lim_{n \to \infty} \sqrt{1 + \frac1n} \\
\Rightarrow && \lim_{n \to \infty} \left (1 + \frac1n \right)^n && \leq e &\leq & \lim_{n \to \infty} \left (1 + \frac1n \right)^{n} \\
\end{align*}
And therefore equality must hold.
This was quite a popular question, as candidates seemed to like using the log series, and appreciated the helpful structuring of the question. However, inequalities are seldom entirely confidently handled, and explanations (wherever they are required) are generally rather feeble. Thus, several marks were often not picked up, sometimes because the candidates did not think they had to consider addressing issues such as whether the series was valid in this case. Part (i) was usually fairly well done, with (ii) providing more of a challenge. Part (iii) required only informal arguments, but many scored only 1 of the 2 marks allocated here due to being a bit too vague about what was going on.