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?

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

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.



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


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.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.