show $ \ln(1+\frac{1}{x})−\frac{1}{x+1}>0 $

Question

show $ \ln(1+\frac{1}{x})−\frac{1}{x+1}>0 $ $;$ $\forall x>1$. I tried to show $ \frac{1}{2x^2} - \frac{2}{3x^3} + \frac{3}{4x^4} - \frac{4}{5x^5} \pm ... >0$ $;$ $\forall x > 1$ . help me.

Answer 1

If $1 \le t \le 1 + \dfrac 1x$ then $\dfrac{x}{x+1} \le \dfrac 1 t \le 1$. Thus $$ \log(1 + \frac 1x) = \int_1^{1 + \frac 1x} \frac 1t \, dt \ge \frac{x}{x+1} \int_1^{1+\frac 1x} \, dt = \frac 1{x+1}.$$

Answer 2

hint

It is equivalent to prove that

for $x>1$,

$$(x+1)\ln(1+\frac 1x)>1$$ or

$$(1+\frac 1x)\ln(1+\frac 1x)>\frac 1x$$

Answer 3

If you want to do this by power series, it's easier to first show that $\ln(1+x) < x$ for all $x\in (-1,1)$, then use a change of variables to get back to your inequality.

Answer 4

Using Lagrange's MVT for $\ln(1+t)$ in $[0,\frac1x]$, you have $$ \ln(1+\frac1x)-\ln1=\frac1{1+c}\cdot\frac1x $$ for some $c\in(0,\frac1x)$. Since $c<\frac1x$, you have $$ \frac1{1+c}>\frac1{1+\frac1x}=\frac{x}{x+1}$$ and hence $$ \ln(1+\frac1x)=\frac1{1+c}\cdot\frac1x>\frac{1}{x+1}. $$