Taylor's Theorem and natural logarithmic inequality Use Taylor's theorem to prove that, for $x>0$.
$$
\ln x+\frac{1}{x}-\frac{1}{2 x^{2}}<\ln (x+1)<\ln x+\frac{1}{x}
$$
The RHS is indeed obvious: algebraic manipulations yield to show $e^{\frac{1}{x}} > 1 + \frac{1}{x}$ which is clear from the first-degree Taylor polynomial of $e^x$ around $x_{0} = 0$. However, after several substitutions, algebraic approaches, and Taylor expansions, similar results for the LHS could not be resolved. Are there alternative ideas to come up for the left-part?
 A: The Taylor’s expansion of the function $\;\ln(1+z)\;$ at $\;z=0\;$ is
$\ln(1+z)=z-\dfrac{z^2}2+\dfrac{z^3}3-\dfrac{z^4}4+\ldots+(-1)^n\dfrac{z^n}n+\ldots$
for all $\;z\in\big]-1,1\big]\;.$
Hence,
$z-\dfrac{z^2}2<\ln(1+z)\quad$ for all $\;z>0\;\land\;z\leqslant1\;.$
By letting $\;z=\dfrac1x\;,\;$ it follows that
$\dfrac1x-\dfrac1{2x^2}<\ln\left(1+\dfrac1x\right)\quad$ for all $\;x\geqslant1\;,$
$\dfrac1x-\dfrac1{2x^2}<\ln\left(\dfrac{x+1}x\right)\quad$ for all $\;x\geqslant1\;,$
$\dfrac1x-\dfrac1{2x^2}<\ln(x+1)-\ln x\quad$ for all $\;x\geqslant1\;,$
$\color{blue}{\ln x+\dfrac1x-\dfrac1{2x^2}<\ln(x+1)\quad\text{ for all }\;x\geqslant1}\;.$
For all $\;x\in\big]0,1\big[\;,\;$ it results that
$\ln x+\dfrac1x-\dfrac1{2x^2}<\ln x+\dfrac12\left[1-\left(\dfrac1x-1\right)^2\right]<$
$<\ln x+\dfrac12<\ln x+\ln2<\ln(2x)<\ln(x+1)\;\;.$
Hence,
$\color{blue}{\ln x+\dfrac1x-\dfrac1{2x^2}<\ln(x+1)\quad\text{ for all }\;x\in\big]0,1\big[}\;.$
Since the Taylor’s expansion of the function $\;e^z\;$ at $\;z=0\;$ is
$e^z=1+z+\dfrac{z^2}{2!}+\dfrac{z^3}{3!}+\ldots+\dfrac{z^n}{n!}+\ldots\;$ for all $\;z\in\mathbb{R}\;\;,$
it follows that
$1+z<e^z\quad$ for all $\;z>0\;,$
$\ln(1+z)<z\quad$ for all $\;z>0\;.$
By letting $\;z=\dfrac1x\;,\;$ we get that
$\ln\left(1+\dfrac1x\right)<\dfrac1x\quad$ for all $\;x>0\;,$
$\ln\left(\dfrac{x+1}x\right)<\dfrac1x\quad$ for all $\;x>0\;,$
$\ln(x+1)-\ln x<\dfrac1x\quad$ for all $\;x>0\;,$
$\color{blue}{\ln(x+1)<\ln x+\dfrac1x\quad\text {for all }\;x>0}\;.$
So far we have proved that
$\color{blue}{\ln x+\dfrac1x-\dfrac1{2x^2}<\ln(x+1)<\ln x+\dfrac1x\quad\text{ for all }\;x>0}\;.$



Addendum :
Gary proved that
$\displaystyle\ln(1+y)=y-y^2\int_0^1\!\dfrac t{1+yt}\,\mathrm{d}t\quad$ for all $\;y>0\;.$
Since $\displaystyle\;0<\int_0^1\!\dfrac t{1+yt}\,\mathrm{d}t<\int_0^1\!t\,\mathrm{d}t=\dfrac12\quad$ for all$\;y>0\;,\;$
it follows that
$y-\dfrac{y^2}2<\ln(1+y)<y\quad$ for all $\;y>0\;.$
A: Hint: Use the Taylor theorem with the reminder written in Lagrange form.
$$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2+\frac{f'''(\xi)}{6}(x-a)^3$$
for some $\xi\in(a,x)$
A: Let $f:(-1,\infty)$ be the function given by $f(y)=\ln (1+y)$. Then
$$
\begin{aligned}
f(y) &= \ln(1+y)\ ,&f(0)&=0\ ,\\
f'(y) &= \frac 1{1+y}\ ,&f'(0)&=1\ ,\\
f''(y) &= -\frac 1{(1+y)^2}\ ,&f''(0)&=-1\ ,\\
f'''(y) &= +\frac 2{(1+y)^3}>0\ ,
\end{aligned}
$$
and we apply the Taylor formula with remainder,
the Taylor formula with remainder
around $0$ for the point $y=1/x$ for the given $x>0$.
We obtain:
$$
\begin{aligned}
\ln(1+x)-\ln x
&=\ln\left(1+\frac 1x\right)\\
&=\ln(1+y)
\\
&=f(y)
\\
&=f(0)+\frac 1{1!}f'(0)\;y+\frac 1{2!}f''(0)\;y^2+\frac 1{3!}\underbrace{f'''(\xi)}_{>0}\;\underbrace{y^3}_{>0}
\\
&\qquad\text{ (for some intermediate point $\xi$ between $0$ and $y$)}
\\
&>f(0)+\frac 1{1!}f'(0)\;y+\frac 1{2!}f''(0)\;y^2
\\
&=0+y-\frac 12y^2
\\
&=0+\frac 1x-\frac 1{2x^2}
\ .
\end{aligned}
$$
