# Math Help - Proving this using Maclaurin's.

1. ## Proving this using Maclaurin's.

Can someone show me how you'd prove:

2. First, can you tell me what is the MacLaurin series for ln(x+1)?

3. For any $0\not = x>-1$ we can find $y$ between $0$ and $x$ so that:
$\log (1+x) - x = \frac{f''(y)}{2!}x^2 < 0$
Since the second derivative is negative.

Therefore, $\log (1+x) < x$ for $-1.

4. Originally Posted by ThePerfectHacker
For any $0\not = x>-1$ we can find $y$ between $0$ and $x$ so that:
$\log (1+x) - x = \frac{f''(y)}{2!}x^2 < 0$
Since the second derivative is negative.

Therefore, $\log (1+x) < x$ for $-1.
I think it's important to say that

$f(x) = \ln (x+1)$

Obvious to most but not all

5. Another proof:

For each $x>-1$ it's $\ln (x+1)=\int_{1}^{x+1}{\frac{dy}{y}}.$ Hence $\frac{1}{x+1}\le \frac{1}{y}\le 1\implies \frac{x}{x+1}\le \ln (x+1)\le x.\quad\blacksquare$