Year: 2012
Paper: 3
Question Number: 4
Course: UFM Pure
Section: Taylor series
The number of candidates attempting more than six questions was, as last year, about 25%, though most of these extra attempts achieved little credit.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item
Show that
\[
\sum_{n=1} ^\infty
\frac{n+1}{n!}
= 2\e - 1
\]
and
\[
\sum _{n=1}^\infty
\frac {(n+1)^2}{n!} = 5\e-1\,.
\]
Sum the series $\displaystyle
\sum _{n=1}^\infty
\frac {(2n-1)^3}{n!}
\,.$
\item
Sum the series
$\displaystyle
\sum_{n=0}^\infty \frac{(n^2+1)2^{-n}}{(n+1)(n+2)}\,$,
giving your answer in terms of natural logarithms.
\end{questionparts}\begin{questionparts}
\item
\begin{align*}
\sum_{n=1}^{\infty} \frac{n+1}{n!} &= \sum_{n=1}^\infty \left ( \frac{1}{(n-1)!} + \frac{1}{n!} \right) \\
&= \sum_{n=0}^\infty \frac{1}{n!} + \sum_{n=1}^\infty \frac{1}{n!} \\
&= \sum_{n=0}^\infty \frac{1}{n!} + \sum_{n=0}^\infty \frac{1}{n!} - 1 \\
&= e + e - 1 \\
&= 2e-1
\end{align*}
\begin{align*}
\sum_{n=1}^{\infty} \frac{(n+1)^2}{n!} &= \sum_{n=1}^{\infty} \frac{n(n-1) + 3n + 1}{n!} \\
&= \sum_{n=2}^{\infty} \frac{1}{(n-2)!} + 3 \sum_{n=1}^\infty \frac1{(n-1)!} + \sum_{n=1}^\infty \frac{1}{n!} \\
&= \sum_{n=0}^{\infty} \frac{1}{n!} + 3 \sum_{n=0}^\infty \frac1{n!} + \sum_{n=0}^\infty \frac{1}{n!} -1 \\
&= 5e-1
\end{align*}
\begin{align*}
\sum_{n=1}^\infty \frac{(2n-1)^3}{n!} &= \sum_{n=1}^\infty \frac{8n^3-12n^2+6n-1}{n!} \\
&= \sum_{n=1}^\infty \frac{8n(n-1)(n-2)+12n^2-10n-1}{n!} \\
&= \sum_{n=1}^\infty \frac{8n(n-1)(n-2)+12n(n-1)+2n-1}{n!} \\
&= 8 e+12e+2e-(e-1) \\
&=21e+1
\end{align*}
\item \begin{align*}
\frac{n^2+1}{(n+1)(n+2)} &= \frac{n^2+3n+2-3n-1}{(n+1)(n+2)}\\
&= 1 - \frac{3n+1}{(n+1)(n+2)} \\
&= 1 + \frac{2}{n+1} - \frac{5}{n+2}
\\
-\log(1-x) &= \sum_{n=1}^\infty \frac1{n}x^{n} \\
\log(2) &= \sum_{n=1}^\infty \frac{2^{-n}}{n} \\
\sum_{n=0}^\infty \frac{(n^2+1)2^{-n}}{(n+1)(n+2)} &= \sum_{n=0}^{\infty} 2^{-n} + 2 \sum_{n=0}^\infty \frac{2^{-n}}{n+1}-5 \sum_{n=0}^\infty \frac{2^{-n}}{n+2} \\
&= 2 + 2\log2-5 \sum_{n=2}^\infty \frac{2^{-n+2}}{n} \\
&= 2 + 2 \log 2 - 5 \left (2\log 2 - 2 \right) \\
&= 12-8\log2
\end{align*}
\end{questionparts}
Just over 70% of the candidates attempted this, with marginally less success than question 3. Lots of attempts relied on manipulating series for e, and would have struggled had the first two results not been given, and even so, there were varying levels of success and conviction. This approach fell apart in the third part with the cubic term. Some candidates used a generating function method successfully with an approach. However, whilst this worked well for part (i), it got very nasty for part (ii). There were lots of sign errors with the log series in part (ii), having begun well with partial fractions.