Natural log question

Question

Prove that for $x> 1$, $\ln(x+1) - \ln(x)< \frac{2}{x+1}$.

Answer 1

Hint:

You could denote a function $g(x) = \ln (x+1)-\ln x-2/(x+1)$, then you can check the $g^{'}(x)$ to see whether $g(x)$ increases or not when $x>1$, then you check $g(1)$ to see whether the inequality meets or not.

Answer 2

Herein, we present a way forward that relies on pre-calculus tools only. To that end, we proceed with a primer.

PRIMER:

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequality

$$\bbox[5px,border:2px solid #C0A000]{\log(x)\le x-1} \tag 1$$

for $x>0$.

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Applying $(1)$, we see that for $1< x$

$$\begin{align} \log(x+1)-\log(x)&=\log\left(1+\frac1x\right)\\\\ &\le \frac1x\\\\ &< \frac2{x+1} \end{align}$$

And we are done!

Answer 3

Let $f(x)=\frac{2}{x+1}-\ln\left(1+\frac{1}{x}\right)$ on $(1,+\infty)$.

Hence, $f'(x)=-\frac{2}{(x+1)^2}+\frac{\frac{1}{x^2}}{1+\frac{1}{x}}=\frac{1-x}{x(x+1)^2}<0$, which says that $f$ is a decreasing function

and since $\lim\limits_{x\rightarrow+\infty}f(x)=0$, we are done!