Defintion of e as a limit

[blerg]

Homework Statement

Show that $$x_n = \left(1 + \frac{1}{n}\right)^{n+1}$$ is decreasing and bounded below

Homework Equations

The Attempt at a Solution

I have tried all sorts of things and none some to be working. I tried expanding it using the binomial theorem, but there are more terms in $$x_{n+1}$$ than $$x_n$$ so i didn't see an easy way to compare them. I tried looking at $$\frac{x_{n+1}}{x_n}$$ and $$x_nx_{n+1}$$ with nothings giving me and progress. Any suggestions?

 

[mighty2000]

Dear student,

To show that x_n is decreasing, we can take the derivative of x_n and show that it is always negative. Let's start by rewriting x_n as:

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

Now, taking the natural logarithm of both sides, we get:

\ln(x_n) = \ln\left[\left(\frac{n+1}{n}\right)^{n+1}\right] = (n+1)\ln\left(\frac{n+1}{n}\right)

Next, we can take the derivative of both sides with respect to n:

\frac{d}{dn}\ln(x_n) = \frac{d}{dn}\left[(n+1)\ln\left(\frac{n+1}{n}\right)\right]

Using the product rule and the chain rule, we get:

\frac{1}{x_n}\cdot\frac{dx_n}{dn} = \ln\left(\frac{n+1}{n}\right) + (n+1)\cdot\frac{1}{\frac{n+1}{n}}\cdot\left(\frac{n}{n+1}\right)^2

Simplifying this expression, we get:

\frac{dx_n}{dn} = -\frac{n+1}{n^2}\left[\ln\left(\frac{n+1}{n}\right) - \frac{n}{n+1}\right]

Now, we can see that the expression in the square brackets is always negative, since n > 0. Therefore, the derivative of x_n is always negative, which means that x_n is decreasing.

To show that x_n is bounded below, we can use the fact that e = \lim_{n\to\infty}\left(1 + \frac{1}{n}\right)^n. This means that for any n > 0, we have:

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

Multiplying both sides by \left(1 + \frac{1}{n}\right), we get:

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