Collection of Maths Problems

Sequences going to e

Task number: 2872

• Hint

  Use Bernoulli's inequality.

• Resolution

  Let \(a_n=\left(1+\frac1n\right)^n=\left(\frac{n+1}n\right)^n\).

  \(\frac{a_{n-1}}{a_n}= \frac{\left(\frac{n}{n-1}\right)^{n-1}}{\left(\frac{n+1}n\right)^n} = \frac{n}{n+1}\left(\frac{n^2}{n^2-1}\right)^{n-1} = \frac{n}{n+1}\left(1+\frac{1}{n^2-1}\right)^{n-1} \le \frac{n}{n+1}\left(1+\frac{n-1}{n^2-1}\right)= \frac{n^2+2n}{n^2+2n+1}<1 \).

  So \(a_n>a_{n-1}\) and the sequence is increasing.

  Now let \(b_n=\left(1+\frac1n\right)^{n+1}=\left(\frac{1+n}n\right)^{n+1}\).

  \(\frac{b_n}{b_{n-1}}= \frac{\left(\frac{n+1}n\right)^{n+1}}{\left(\frac{n}{n-1}\right)^{n}} = \frac{n+1}{n}\left(\frac{n^2-1}{n^2}\right)^{n} = \frac{n+1}{n}\left(1-\frac{1}{n^2}\right)^{n} \le \frac{n}{n+1}\left(1-\frac{n}{n^2}\right)= \frac{n-1}{n+1}<1. \)

  So \(b_n<b_{n-1}\) and the sequence is decreasing.

  Because \(a_n=\left(1+\frac1n\right) b_n\) we know that \(a_n<b_n\). From the monotonicity of both sequences we can deduce that in fact \(a_n<b_{n'}\) for any \(n,n'\in\mathbb N\).

  It remains to show that \(e=\inf\{a_n,n\in \mathbb N\}\) is the limit of the sequences \(a_n\) and \(b_n\). For all \(n\) we know that \(a_n > e > b_n\). What is more, \(\displaystyle \lim_{n\to\infty} \frac{a_n}{b_n}= \lim_{n\to\infty} \left(1+\frac1n\right) =1 \).

  Of course also \(\displaystyle \lim_{n\to\infty} {a_n}-{b_n}= \lim_{n\to\infty} b_n\frac{{a_n}-{b_n}}{b_n}\le 4 \lim_{n\to\infty} \frac{{a_n}-{b_n}}{b_n}= 4 \left(\lim_{n\to\infty} \frac{a_n}{b_n}-1\right)=0 \).

  So for any \(\varepsilon\) we can find \(n_0\) such that \(a_n-b_n<\varepsilon\) for any \(n\ge n_0\). It follows directly that \(a_n-e<\varepsilon\) and \(e-b_n<\varepsilon\).