A Peculiar Infinite Limit!

Calculus Level 4

L = lim n [ ( n + 1 ) ! ( e x n + 1 e x n ) ] \large{L=\lim_{n \to \infty} \left[ (n+1)! \ (e^{x_{n+1}} - e^{x_{n}}) \right ] }

If x n = k = 0 n 1 k ! {x_n = \displaystyle \sum_{k=0}^n \dfrac{1}{k!}} , then find the value of ln ( L ) {\ln(L)} . Give your answer to 3 decimal places.


The answer is 2.718.

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1 solution

Satyajit Mohanty
Jul 20, 2015

L = lim n [ ( n + 1 ) ! ( e x n + 1 e x n ) ] L= \lim_{n \to \infty} \left[ (n+1)! (e^{x_{n+1}} - e^{x_n})\right]

= lim n [ ( n + 1 ) ! e x n ( e x n + 1 x n 1 ) ] = \lim_{n \to \infty} \left[ (n+1)! \cdot e^{x_n}(e^{x_{n+1} - x_n} - 1)\right]

= lim n [ ( n + 1 ) ! e x n ( e 1 ( n + 1 ) ! 1 ) ] = \lim_{n \to \infty} \left[ (n+1)! \cdot e^{x_n}\left (e^{\frac{1}{(n+1)!}} - 1 \right)\right]

= lim n e x n × lim n ( e 1 ( n + 1 ) ! 1 1 ( n + 1 ) ! ) = \lim_{n \to \infty} e^{x_n} \times \lim_{n \to \infty} \left( \frac{e^{\frac{1}{(n+1)!}} - 1}{\frac{1}{(n+1)!}}\right)

= e e × 1 = e e = e^e \times 1 = e^e

Therefore, ln ( L ) = ln ( e e ) = e 2.718 \ln(L) = \ln(e^e) = e \approx \boxed{2.718} .

Is this correct? The first limit in your answer does not tend to 1 and the equation in the second line is not true.

Felipe Álvares - 5 years, 10 months ago

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I've updated the solution. Please check this :)

Satyajit Mohanty - 5 years, 10 months ago

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Much better!

Felipe Álvares - 5 years, 10 months ago

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