2022HSCShort answer

Q14 (b) (v)

1 mark

Question

Let $J_n=\int_0^1x^ne^{-x}\,dx$ . Using parts (ii) and (iv), prove that
$e=\lim_{n\to\infty}\sum_{r=0}^{n}\frac1{r!}.$

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	1

Sample answer:
From part (ii),
$0\le J_n\le\frac1{n+1}$
But
$\lim_{n\to\infty}\frac1{n+1}=0$
So $\lim_{n\to\infty}J_n=0$ .

From part (iv),
$\frac{J_n}{n!}=1-\frac1e\sum_{r=0}^{n}\frac1{r!}$
$\sum_{r=0}^{n}\frac1{r!}=e-\frac{eJ_n}{n!}$

\lim_{n\to\infty}\left(\sum_{r=0}^{n}\frac1{r!}\right) =\lim_{n\to\infty}\left(e-\frac{eJ_n}{n!}\right)=e-0=e

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Syllabus alignment

ProofCalculus

The Nature of ProofUse formal proof language, contradiction, counter-examples and rigorous arguments involving numbers and inequalities.
Further IntegrationApply substitution, partial fractions, integration by parts and recurrence relationships to theoretical and practical problems.

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