Year: 2017
Paper: 3
Question Number: 1
Course: UFM Pure
Section: Sequences and series, recurrence and convergence
The total entry was only very slightly smaller than that of 2016, which was a record entry, but was still over 10% more than 2015. No question was attempted by in excess of 90%, although two were very popular and also five others were attempted by 60% or more. No question was generally avoided with even the least popular one attracting more than 10% of the entry. Less than 10% of candidates attempted more than 7 questions, and, apart from 18 exceptions, those doing so did not achieve very good totals and seemed to be 'casting around' to find things they could do: the 18 exceptions were very strong candidates who were generally achieving close to full marks on all the questions they attempted. The general trend was that those with six attempts fared better than those with more than six.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
\begin{questionparts}
\item
Prove that, for any positive integers $n$ and $r$,
\[
\frac{1}{^{n+r}\C_{r+1}}
=\frac{r+1}{r} \left(\frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}}\right).
\]
Hence determine
\[
\sum_{n=1}^{\infty}{\frac{1}{^{n+r}\C_{r+1}}}
\,,
\]
and deduce that \
$\displaystyle \sum_{n=2}^\infty \frac 1 {^{n+2}\C_3} = \frac12\,$.
\item Show that,
for $n \ge 3\,$,
\[
\frac{3!}{n^3} < \frac{1}{^{n+1}\C_{3}}
\ \ \ \ \
\text{and}
\ \ \ \ \
\frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}}
< \frac{5!}{n^3}
\,.
\]
By summing these inequalities for $n \ge 3\,$, show that
\[
\frac{115}{96} < \sum_{n=1}^{\infty}{\frac{1}{n^3}} <
\frac{116}{96} \, .
\]
\end{questionparts}
{\bf Note: } $^n\C_r$ is another notation for $\displaystyle \binom n r $.\begin{align*}
\frac{r+1}{r} \left(\frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}}\right) &= \frac{r+1}{r} \l \frac{r!(n-1)!}{(n+r-1)!} - \frac{r!n!}{(n+r)!} \r \\
&= \frac{(r+1)!(n-1)!}{r(n+r-1)!} \l 1 - \frac{n}{n+r} \r \\
&= \frac{(r+1)!(n-1)!}{r(n+r-1)!} \frac{r}{n+r} \\
&= \frac{(r+1)!n!}{(n+r)!} \\
&= \frac{1}{^{n+r}\C_{r+1}}
\end{align*}
\begin{align*}
\sum_{n=1}^{\infty}{\frac{1}{^{n+r}\C_{r+1}}} &= \sum_{n=1}^{\infty} \l \frac{r+1}{r} \left(\frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}}\right) \r \\
&= \frac{r+1}{r} \sum_{n=1}^{\infty} \l \frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}} \r \\
&= \frac{r+1}{r} \lim_{N \to \infty} \sum_{n=1}^{N} \l \frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}} \r \\
&= \frac{r+1}{r} \lim_{N \to \infty} \l \frac{1}{^{1+r-1}\C_{r}} - \frac{1}{^{N+r}\C_{r}}\r \\
&= \frac{r+1}{r} \frac{1}{^{1+r-1}\C_{r}} \tag{since $\frac{1}{^{N+r}\C_{r}} \to 0$} \\
&= \frac{r+1}{r}
\end{align*}
When $r = 2$, we have:
\begin{align*}
&& \frac{3}{2} &= \sum_{n=1}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\
&& &=\frac{1}{^{1+2}\C_{3}} + \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\
&& &= 1 + \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\
\Rightarrow && \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} &= \frac12
\end{align*}
\begin{align*}
\frac{1}{^{n+1}\C_{3}} &= \frac{3!}{(n+1)n(n-1)} \\
&= \frac{3!}{n^3-n} \\
&> \frac{3!}{n^3}
\end{align*}
\begin{align*}
\frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} &= \frac{5!}{(n+1)n(n-1)} - \frac{5!}{(n+2)(n+1)n(n-1)(n-2)} \\
&= \frac{5!}{n^3} \frac{n^2}{n^2-1}\l 1- \frac{1}{n^2-4} \r \\
&= \frac{5!}{n^3} \frac{n^2}{n^2-1}\l \frac{n^2-5}{n^2-4} \r \\
&= \frac{5!}{n^3} \frac{n^2(n^2-5)}{(n^2-1)(n^2-4)} \\
&< \frac{5!}{n^3}
\end{align*}
Since $k(k-5) < (k-1)(k-4) \Leftrightarrow 0 < 4$, this only makes sense if $n \geq 3$
\begin{align*}
&&\frac{3!}{n^3} &< \frac{1}{^{n+1}\C_{3}} \tag{if $n \geq 3$} \\
\Rightarrow &&\sum_{n=3}^\infty \frac{3!}{n^3} &< \sum_{n=3}^\infty \frac{1}{^{n+1}\C_{3}} \\
\Rightarrow && \frac{6}{1^3} + \frac{6}{2^3} + \sum_{n=3}^\infty \frac{3!}{n^3} &< \frac{6}{1^3} + \frac{6}{2^3} + \sum_{n=3}^\infty \frac{1}{^{n+1}\C_{3}} \\
\Rightarrow && \sum_{n=1}^\infty \frac{3!}{n^3} &< 6 + \frac{3}{4} + \sum_{n=2}^\infty \frac{1}{^{n+2}\C_{2+1}} \\
\Rightarrow && \sum_{n=1}^\infty \frac{3!}{n^3} &< 6 + \frac{3}{4} + \frac{1}{2} = \frac{29}{4} \\
\Rightarrow && \sum_{n=1}^\infty \frac{1}{n^3} &< \frac{29}{24} = \frac{116}{96} \\
\end{align*}
\begin{align*}
&& \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} &< \frac{5!}{n^3} \\
\Rightarrow && \sum_{n=3}^\infty \l \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} \r &< \sum_{n=3}^\infty \frac{5!}{n^3} \\
\Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{20}{^{n+1}\C_3} - \sum_{n=3}^\infty \frac{1}{^{n+2}\C_{5}} &< \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{5!}{n^3} \\
\Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=2}^\infty \frac{20}{^{n+2}\C_{2+1}} - \sum_{n=1}^\infty \frac{1}{^{n+4}\C_{4+1}} &< \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{5!}{n^3} \\
\Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \frac{20}{2} - \frac{4+1}{4} &< \sum_{n=1}^\infty \frac{5!}{n^3} \\
\Rightarrow && \frac{115}{96} &< \sum_{n=1}^\infty \frac{1}{n^3} \\
\end{align*}
The most popular question on the paper, attempted by about 84% of the candidates, it was also the most successfully answered with an average mark of about 12/20. The first result was generally well answered with a few candidates attempting to use induction, and then proving the result directly. The summations were usually done well, though often lacked explanation. Usually, the inequalities were not well argued, there was poor layout, and no mention of positivity. Those who spotted the link with part (i) did well in general summing the inequalities, though there were some problems with the indices.