Prove that $e^x > 1 + (1 + x)\log(1 + x), x > 0$ using power series expansion. Prove that $e^x > 1 + (1 + x)\log(1 + x), x > 0$ using power series expansion.
I am a bit puzzled by this statement because the power series for $\log(1+x)$ only converges iff $|x|<1$. Is the problem sound? Should it be $1>x>0$?
 A: We can make an estimate for $1+(1+x)\log(1+x)$ that does not rely on the power series for $\log(1+x)$, and instead use the power series for $e^x$, which is valid for $x\ge0$.
Since $\frac1{1+t}\le1$ on $[0,x]$,
$$
\begin{align}
\log(1+x)
&=\int_0^x\frac{\mathrm{d}t}{1+t}\tag1\\
&\le x\tag2
\end{align}
$$
Furthermore,
$$
\begin{align}
(1+x)\log(1+x)-x
&=x\cdot\frac1x\int_0^x\log(1+t)\,\mathrm{d}t\tag3\\
&\le x\log\left(1+\frac1x\int_0^xt\,\mathrm{d}t\right)\tag4\\[3pt]
&=x\log(1+x/2)\tag5\\[3pt]
&\le\frac{x^2}2\tag6
\end{align}
$$
Explanation:
$(4)$: since $\log(1+t)$ is concave, apply Jensen's Inequality
$(6)$: apply $(2)$
Therefore, by comparing to the Taylor series for $e^x$,
$$
\begin{align}
1+(1+x)\log(1+x)
&\le1+x+\frac{x^2}2\tag7\\
&\le e^x\tag8
\end{align}
$$
Explanation:
$(7)$: apply $(6)$
$(8)$: $e^x=1+x+\frac{x^2}2+\text{positive terms}$

A: hint
Let $$f(x)=e^x-1-(x+1)\ln(1+x)$$
For $ x>0 $, $ f $ is twice differentiable at $ [0,x] $, thus by Taylor-Lagrange formula, there exists $ c\in (0,x) $ such that
$$f(x)=f(0)+xf'(0)+\frac{x^2}{2}f''(c)$$
$$=\frac{x^2}{2}\frac{(1+c)e^c-1}{1+c}$$
A: There are multiple Taylor expansions of a function, you have chosen the expansion about the origin. Consider that any Taylor expansion of $\log(1+x)$ will have the same radius of convergence of $\frac{1}{1+x}$. Well
$$\frac{1}{1 + x} = \frac{1}{1+c -(c-x)} = \Big(\frac{1}{1+c}\Big) \frac{1}{1 -(\frac{c-x}{1+c})} = \frac{1}{1+c}\sum_{n=0}^\infty\Big(\frac{c-x}{1+c}\Big)^n $$
The series above converges as long as $|x-c| < |1+c|$ and $c\not=-1$. In effect this means that you can find an expansion anywhere as long as it is away from $c=-1$. In particular you have that
$$ \log(1 + x) = \log(1+c) + \frac{1}{1+c}\sum_{n=0}^\infty \frac{1}{n+1}\Big(\frac{c-x}{1+c}\Big)^{n+1} $$
Since your asks about any $x>0$, you can use this approach by first fixing an $x$ and choosing an expansion to meet your needs.
A: Let
$$ f(t)=e^t-1, g(t)=(t+1)\log(t+1). $$
Just simply usig Cauchy's MVT, one has, for $x>0$
$$ \frac{f(x)-f(0)}{g(x)-g(0)}=\frac{f'(c)}{g'(c)} $$
where $c\in(0,x)$. Hence
$$ \frac{e^x-1}{(x+1)\log(x+1)}=\frac{e^c}{\log(c+1)+1}. \tag1$$
Again using Cauchy's MVT, one has
$$ \frac{e^c-1}{\log(1+c)}=(c_1+1)e^{c_1}>1, c_1\in(0,c)$$
which gives
$$ e^c>\log(c+1)+1. $$
So one has
$$ e^x>1+(x+1)\log(x+1). $$
A: For $0<x<1$, does the following proof work?
$e^x>1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}$ and $\log(1+x)<x-\dfrac{x^2}{2}+\dfrac{x^3}{3}$ and so,
$$1+(1+x)\log(1+x)<1+x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{3}$$
We observe that $$1+x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{3}<1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}$$
iff
$x^3>x^4$ which is true for $x\in (0,1)$.
