June 2023 Paper 3 Q14
14
The questions in this section refer to the article on the Insert. You should read the article before attempting the questions.
The relevant parts of the article “Approximating series” are reproduced below; the line numbers are those printed on the Insert.
Line 31
Applying Euler’s approximate summation formula to the harmonic seriesLines 32–33
Using Euler’s approximate summation for the harmonic series gives
\(\displaystyle\sum_{r=1}^{n}\frac{1}{r} \approx \int_1^n \frac{1}{x}\,\mathrm{d}x + \frac{1}{2}\left(\frac{1}{n} + 1\right) + \frac{1}{12}\left(1 - \frac{1}{2}\right) - \frac{1}{12}\left(\frac{1}{n} - \frac{1}{n+1}\right)\).Line 34
This simplifies to \(\displaystyle\sum_{r=1}^{n}\frac{1}{r} \approx \ln n + \frac{13}{24} + \frac{6n+5}{12n(n+1)}\).
Show that the expression given in line 33 simplifies to \(\displaystyle\sum_{r=1}^{n}\frac{1}{r} \approx \ln n + \frac{13}{24} + \frac{6n+5}{12n(n+1)}\), as given in line 34. [3]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{1}{2n} + \dfrac{1}{2} + \dfrac{1}{12} - \dfrac{1}{24} - \dfrac{1}{12n} + \dfrac{1}{12(n+1)}\) | M1 | 1.1a |
| \(\left[\displaystyle\int_1^n \frac{1}{x}\,\mathrm{d}x\right] = \left[\ln x\right]_1^n = \ln n - \ln 1 = \ln n\) | B1 | 1.1 |
| \(\ln n + \dfrac{13}{24} + \dfrac{6(n+1) - 1}{12n(n+1)} = \ln n + \dfrac{13}{24} + \dfrac{6n+5}{12n(n+1)}\) | A1 | 2.2a |
| [3] |
Notes
M1: Or better
B1: Integral correctly evaluated
ln 1 or “\(\ln n - 0\)” seen.
A1: AG. Convincing completion