P-n-e-1-1-1-2-1-3-1-4-1-n-1-1-n-e-1-1-2-1-3-1-4-1-n-1-1-n-e-k-1-n-1-k-1-k-P-l

Question Number 175770 by mnjuly1970 last updated on 06/Sep/22

P_{n} = e^{(\frac{1}{1} - \frac{1}{2}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{n - 1} - \frac{1}{n})}

= e^{(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots + \frac{1}{n - 1} - \frac{1}{n})}

= e^{\underset{k = 1}{\sum^{n}} \frac{{(- 1)}^{k + 1}}{k}}

∴ P = {l i m}_{n \to \infty} (e^{\underset{k = 1}{\sum^{n}} \frac{{(- 1)}^{k + 1}}{k}})

= e^{{l i m}_{n \to \infty} (\underset{k = 1}{\sum^{n}} \frac{{(- 1)}^{k + 1}}{k})}

= e^{l n (2)} = 2