Show that $\sin(x) > \ln(x+1)$ for any $x \in (0,1)$ Show that $\sin(x) > \ln(x+1)$ when $x \in (0,1)$. 
I'm expected to use the maclaurin series (taylor series when a=0)
So if i understand it correctly I need to show that: 
$$\sin(x) = \lim\limits_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{(-1)^{k-1}}{(2k-1)!} \cdot x^{2k-1} > \lim\limits_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} \cdot x^k = \ln(x+1)$$
I tried to show that for any k the general term in the bigger sum is greater then the other one (the general term in the smaller sum) but its not true :(.
$$\frac{(-1)^{k-1} \cdot x^{2k-1}}{(2k-1)!} > \frac{(-1)^{k-1} \cdot x^k}{k}$$
when k is odd we get:
$$\frac{x^{k-1}}{(2k-1)!} > \frac{1}{k}$$
and this is a contradiction since :
$x^{k-1} < 1$ for any $x \in (0,1)$ and $k > 1$ and $(2k-1)! > k $ 
so for any $k > 1 $ its $\frac{x^{k-1}}{(2k-1)!} < \frac{1}{k}$ and if $k=1$ its $\frac{x^{k-1}}{(2k-1)!} = \frac{1}{k}$.
What am I doing wrong and how i'm supposed to prove it ? 
Thanks in advance for help .  

Let $f(x)=\sin x-\ln(x+1)$. we try to show that $f(x)$ is increasing. In fact
$$ f'(x)=\cos x-\frac{1}{x+1}=\frac{(x+1)\cos x-1}{x+1}. $$
Now we show $(x+1)\cos x>1$ or $\sec x-1<x$ for $0<x<1$.
Note
\begin{eqnarray}
\sec x-1=\frac{1-\cos x}{\cos x}=\frac{2\sin^2\frac{x}{2}}{1-2\sin^2\frac{x}{2}}.
\end{eqnarray}
Since $\frac{2u}{1-2u}$ is increasing and $\sin x\le x$ for $x\in[0,\pi/2]$, one has
\begin{eqnarray}
\sec x-1=\frac{2\sin^2\frac{x}{2}}{1-2\sin^2\frac{x}{2}}\le \frac{x^2}{2-x^2}\le x^2<x.
\end{eqnarray}
Done.
