Year: 2015
Paper: 2
Question Number: 1
Course: UFM Pure
Section: Taylor series
As in previous years the Pure questions were the most popular of the paper with questions 1, 2 and 6 the most popular. The least popular questions on the paper were questions 8, 11 and 13 with fewer than 250 attempts for each of them. There were many examples of solutions in this paper that were insufficiently well explained given that the answer to be reached had been provided in the question.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
\begin{questionparts}
\item By use of calculus, show that $x- \ln(1+x)$ is positive for all positive $x$. Use this result to show that
\[
\sum_{k=1}^n \frac 1 k > \ln (n+1)
\,.
\]
\item By considering $ x+\ln (1-x)$, show that
\[
\sum_{k=1}^\infty \frac 1 {k^2} <1+ \ln 2
\,.
\]
\end{questionparts}\begin{questionparts}
\item Consider $f(x) = x - \ln (1+ x)$, then $f'(x) = 1 - \frac{1}{1+x} = \frac{x}{1+x} > 0$ if $x >0$.
Therefore $f(x)$ is strictly increasing on the positive reals. Since $f(0) = 0$ we must have $f(x) > 0$ for all positive $x$, ie $x - \ln(1+x)$ is positive for all positive $x$.
\begin{align*}
\sum_{k=1}^n \frac1k &\underbrace{>}_{x > \ln(1+x)} \sum_{k=1}^n \ln \left (1 + \frac1k \right ) \\
&= \sum_{k=1}^n \ln \left ( \frac{k+1}{k} \right ) \\
&= \sum_{k=1}^n \left ( \ln (k+1) - \ln (k) \right) \\
&= \ln (n+1) - \ln 1 \\
&= \ln (n+1)
\end{align*}
\item Let $g(x) = x + \ln (1-x)$ ,then $g'(x) = 1 - \frac{1}{1-x} = \frac{-x}{1-x} < 0$ if $0 < x < 1$ and $g(0) = 0$. Therefore $g(x)$ is decreasing and hence negative on $0 < x < 1$, in particular $x < -\ln(1-x) $
\begin{align*}
\sum_{k=2}^n \frac1{k^2} &\underbrace{<}_{x < -\ln(1+x)} \sum_{k=2}^n - \ln \left (1-\frac1{k^2} \right) \\
&= -\sum_{k=2}^n \ln \left ( \frac{k^2-1}{k^2}\right) \\
&= \sum_{k=2}^n \l 2 \ln k - \ln(k-1) - \ln(k+1) \r \\
&= \ln n - \ln(n+1) - \ln 0+\ln 2 \\
&= \ln 2 + \ln \frac{n}{n+1}
\end{align*}
as $n \to \infty$ we must have $\displaystyle \sum_{k=2}^{\infty} \frac1{k^2} < \ln 2$ ie
\[ \sum_{k=1}^\infty \frac 1 {k^2} <1+ \ln 2\]
\end{questionparts}
This was a popular question, but a number of common errors resulted in a relatively low average score for the attempts made. A number of candidates did not appreciate that it is necessary in the first part to show both that the gradient is positive for all relevant values of x and to check the value when x=0. Additionally, many candidates failed to note that the next part of this question instructed them to use the result shown in the second section of part (i) and instead used a graphical method. Other common errors included an incorrect use of the chain rule in the second part leading to a sign error and incorrect statements of formulae for the sums.