## Sequences going to e

Show that the sequence $$\left(1+\frac1n\right)^n$$ is increasing and the sequence $$\left(1+\frac1n\right)^{n+1}$$ is decreasing.

From there, show that these sequences have the same limit.

• #### 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 $$e-b_n<\varepsilon$$.