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

Therefore, $\displaystyle \log (1+x) < x$ for $\displaystyle -1<x\not = 0$.

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

Therefore, $\displaystyle \log (1+x) < x$ for $\displaystyle -1<x\not = 0$.
I think it's important to say that

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

Obvious to most but not all

5. Another proof:

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