Pre-U
Pre-U 9795
Specimen
Q14
13 marks
Challenging +1.8
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]