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Prove that $\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0$

First inequality using MVT:

$\frac{1}{a+1}<\ln \frac{a+1}{a}:$

$f(a)=\frac{1}{a+1}-\ln \frac{a+1}{a}$

$f(1)=\frac{1-2\ln 2}{2},f^{'}(a)=\frac{1}{a(a+1)^2}>f(1)\Rightarrow f(a)>f(1)$

$\frac{1}{a+1}-\ln \frac{a+1}{a}-\frac{1-2\ln 2}{2}>0$

This is not the starting inequality.

Is there something wrong in this method?

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    $\begingroup$ $\log \frac {a+1}a=\int_a^{a+1}\frac 1 x\,dx$ and $\frac 1 x$ is strictly decreasing. $\endgroup$ – A.S. Dec 29 '15 at 16:54
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METHOD 1: Non-Calculus Based

In This Answer, I showed using basic tools only that the logarithm function satisfies the inequalities

$$\frac{x}{x+1}\le \log(1+x) \le x \tag 1$$

for $x\ge -1$. Note that $\frac{a+1}{a}=1+\frac1a$. Then, setting $x=\frac1a$ in $(1)$ gives the inequalities

$$\frac{1}{a+1}\le \log \left(\frac{a+1}{a}\right)\le \frac1a$$

The strict inequalities follows since the equality in $(1)$ occurs only when $x=0$. Since $\frac1a>0$, we have

$$\frac{1}{a+1} < \log \left(\frac{a+1}{a}\right) < \frac1a$$


METHOD 2: Calculus Based

Form the functions $f(x)=\log \left(1+x\right)-x$ and $g(x)= \log(1+x)-\frac{x}{x+1}$ for $0<x$.

Then, note that $f'(x)=\frac{-1}{1+x}<0$ and $g'(x)= \frac{x}{(x+1)^2}>0$.

Therefore, $f(0)=0$ and $f'(x)<0$ implies $f(x) > 0$ for $x>0$. Then set $x=1/a$ and we have

$$\log \left(\frac{a+1}{a}\right) < \frac1a$$

Finally, $g(0)=0$ and $g'(x)>0$ implies $g(x)>0$. Then, set $x=1/a$ and we have

$$\frac{1}{a+1}< \log \left(\frac{a+1}{a}\right) $$

And we are done.

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HINT:

Take $$f(x)=\ln x $$ and apply MVT in [$a,a+1$].

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For the second inequality multiply by $a$ to get $$\ln \left(1 + \frac 1a\right)^a < 1$$ or $$\left(1 + \frac 1a\right)^a < e $$ which is correct since function $f(a) = \left(1 + \frac 1a\right)^a$ is strictly increasing and ha a limit at $+\infty$ equal to $e$.

To prove the first part use the function $g(a) = \left(1 + \frac 1a\right)^{a+1}$ which converges to $e$ from above.

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Hint: use $f(x)=\ln x$ in $[a,a+1]$.

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