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To prove: $$x-\frac{x^2}{2}<\ln(1+x)<x-\frac{x^2}{2(1+x)},\quad\forall x>0$$

I have used Taylor series expansion at 0 for both the inequalites. The greater than by expanding $\ln(1+x)$ and the less than by expanding $\int \ln(1+x)\,dx$ at 0.

Is there a cleaner / more elegant way of achieving the same?

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Taylor series expansion cannot give you such inequalities, for any fixed x. –  Did Jul 16 '12 at 20:36

2 Answers 2

up vote 1 down vote accepted

Another way to do the lower bound, for example, would be to consider $f(x) = x - \dfrac{x^2}{2} - \ln(1+x)$. $f(0) = 0$.

But also $f'(x) = 1 - x - \dfrac{1}{1+x} = \dfrac{(1-x)(1+x) - 1}{1+x} = \dfrac{1 - x^2 - 1}{1+x} = \dfrac{-x^2}{1+x} < 0$.

So since $f(0) = 0$ and $f' < 0$, $f$ is monotonically and strictly decreasing. Thus $f(x) \leq 0$, and if $x > 0$ we have that $f(x) < 0$. And this says exactly that $x - x^2/2 < \ln(1 + x)$.

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Yes, Yes, absolutely! Thank you, I was looking for this kind of a solution. Much neater. –  Soham Jul 16 '12 at 20:48
    
The Taylor series with either Lagrange or integral form of the remainder is fine. We get that $\ln(1+x)=x-\frac{x^2}{2}+\frac{2}{(1+c)^2}\frac{x^3}{3!}$ for some $c$ between $1$ and $x$, which yields the lower bound. –  André Nicolas Jul 16 '12 at 21:04
    
@AndréNicolas Thanks I will have a look at it. –  Soham Jul 16 '12 at 21:35
    
@Soham: I wrote out a simple argument for the upper bound. One could also use a remainder term argument for that, since remainder term arguments and the more "basic-looking" sign of derivatives arguments are essentially the same. –  André Nicolas Jul 16 '12 at 21:42

The upper bound also yields to standard calculus methods. Equivalently, we want to prove that $(1+x)\ln(1+x) \gt x(1+x)-\frac{x^2}{2}$, or more simply that $x+\frac{x^2}{2} \gt (1+x)\ln(1+x)$.

Let $f(x)=x+\frac{x^2}{2}-(1+x)\ln(1+x)$. We have $f(0)=0$. We will show that $f$ is increasing, and hence posiive, for positive $x$. We have $$f'(x)=1+x-(1+\ln(1+x))=x-\ln(1+x).$$ To show that $f'(x)$ is positive for $x \gt 0$, note that $f'(0)=0$. But $f'(x)$ is increasing, since $f''(x)=1-\frac{1}{1+x} \gt 0$ for $x\gt 0$.

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