Decreasing sequence Anastasia

We want to prove that we have

$$(1+\frac{1}{n})^{n}<e<(1+\frac{1}{n})^{n+1}$$

or equivalent

$$\frac{1}{n+1}<\ln (n+1)-\ln n<\frac{1}{n}$$

We shall prove that sequence

$$x_n=(1+\frac{1}{n})^{n+1}=(\frac{n+1}{n})^{n+1}$$

has the property

$$x_n>x_{n+1}$$

As

$$\lim_{n\to \infty}(1+\frac{1}{n})^{n+1}=e$$

we shall have

$$e<(1+\frac{1}{n})^{n+1}$$

We also know that sequence $ y_n=(1+\frac{1}{n})^{n} $ is increasing and upper bounded and its limit is Euler number e.
So we also have

$$(1+\frac{1}{n})^{n}<e$$

Now we return to sequence $ x_n $ to prove that it is decreasing.
We have to prove

$$x_n=(1+\frac{1}{n})^{n+1}>(1+\frac{1}{n+1})^{n+2}$$

Or

$$x_n=(\frac{n+1}{n})^{n+1}>(\frac{n+2}{n+1})^{n+2}$$

This is equivalent to

$$(\frac{(n+1)^2}{n(n+2)})^{n+1}>\frac{n+2}{n+1}=1+\frac{1}{n+1}$$

or

$$(\frac{n^2+2n+1}{n^2+2n})^{n+1}=(1+\frac{1}{n^2+2n})^{n+1}>1+\frac{1}{n+1}$$

But we have for all a>0

$$(1+a)^n=\sum_{k=0}^{n}C_n^ka^{n-k}>1+na$$
$$(1+\frac{1}{n^2+2n})^{n+1}>1+\frac{n+1}{n^2+2n}$$

So it is enough to prove that

$$1+\frac{n+1}{n^2+2n}>1+\frac{1}{n+1}$$

which is equivalent to

$$\frac{n+1}{n^2+2n}>\frac{1}{n+1}$$

or

$$(n+1)^2>n^2+2n$$

We have also

$$x_n=(1+\frac{1}{n})^{n+1}=(1+\frac{1}{n})^{n}(1+\frac{1}{n})$$

and so

$$ \lim_{n\to \infty}x_n=\lim_{n\to \infty}(1+\frac{1}{n})^{n+1}=\lim_{n\to \infty}(1+\frac{1}{n})^{n}\lim_{n\to \infty}(1+\frac{1}{n})=e$$

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