Challenging +1.8 This is a sophisticated Further Maths integration problem requiring reduction formula derivation via product rule, recursive substitution to find a pattern, and limit analysis. The multi-part structure with 13 marks total, the need to recognize factorial patterns, and the careful justification of convergence place it well above average difficulty, though the guided structure prevents it from being exceptionally hard.
Let \(J_n = \int_1^{\mathrm{e}} (\ln x)^n \, \mathrm{d}x\), where \(n\) is a positive integer. By considering \(\frac{\mathrm{d}}{\mathrm{d}x}(x(\ln x)^n)\), or otherwise, show that
$$J_n = \mathrm{e} - nJ_{n-1}.$$ [4]
Let \(J_n = \frac{J_n}{n!}\). Show that
$$\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \ldots + \frac{1}{10!} = \frac{1}{\mathrm{e}}(1 + J_{10}).$$ [6]
It can be shown that
$$\sum_{r=2}^{n} \frac{(-1)^r}{r!} = \frac{1}{\mathrm{e}}(1 + (-1)^n J_n)$$
for all positive integers \(n\). Deduce the sum to infinity of the series
$$\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \ldots,$$
justifying your conclusion carefully. [3]
Let $J_n = \int_1^{\mathrm{e}} (\ln x)^n \, \mathrm{d}x$, where $n$ is a positive integer. By considering $\frac{\mathrm{d}}{\mathrm{d}x}(x(\ln x)^n)$, or otherwise, show that
$$J_n = \mathrm{e} - nJ_{n-1}.$$ [4]
Let $J_n = \frac{J_n}{n!}$. Show that
$$\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \ldots + \frac{1}{10!} = \frac{1}{\mathrm{e}}(1 + J_{10}).$$ [6]
It can be shown that
$$\sum_{r=2}^{n} \frac{(-1)^r}{r!} = \frac{1}{\mathrm{e}}(1 + (-1)^n J_n)$$
for all positive integers $n$. Deduce the sum to infinity of the series
$$\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \ldots,$$
justifying your conclusion carefully. [3]
\hfill \mbox{\textit{Pre-U Pre-U 9795 Q14 [13]}}