Show that $\frac{1}{1+k}=\frac{\frac{1}{k}}{1+\frac{1}{k}}\leq \ln(1+\frac{1}{k})\leq\frac{1}{k}$

Question

Prove the following: $$\frac{1}{1+k}=\frac{\frac{1}{k}}{1+\frac{1}{k}}\leq \ln(1+\frac{1}{k})\leq\frac{1}{k}$$

I know I can prove it with induction if the values were naturals. However, the "problem" for me is that they're real.

Answer 1

For all $x \in \mathbb{R}$, $$e^x \geq 1 + x$$ Taking log on both sides we get, $$\ln (1 + x) \leq x, \forall x > -1$$ Substituting $x = \frac{1}{k}, k \notin [0, -1]$, we get, $$\displaystyle{\ln \left(1 + \frac{1}{k}\right) \leq \frac{1}{k}}$$ Substituting $x = \frac{-1}{k + 1}, k \notin [0, -1]$, we get, $$\ln \left(\frac{k}{k + 1}\right) \leq \frac{-1}{k + 1}\Rightarrow \ln \left(1 + \frac{1}{k}\right) \geq \frac{1}{k + 1} $$

Answer 2

Set $f:[k,k+1]\to\mathbb{R}$ given by $f(x)=\log(x)$

Then, $f$ is continuous in $[k,k+1]$ and differentiable in $(k,k+1)$. Thus, there is $\xi\in(k,k+1)$ such that $\frac{f(k+1)-f(k)}{k+1-k}=f^\prime(\xi)$.

That is $\log(k+1)-\log(x)=\frac{1}{\xi}$ for some $\xi\in(k,k+1)$. Then

$$\frac{1}{k+1}<\frac{1}{\xi}=\log(k+1)-\log(k)=\log\frac{k+1}{k}=\log(1+\frac{1}{k})<\frac{1}{k}$$

Answer 3

For $x>0$ we have $$\log (1+x)=\int_1^{1+x}\frac {1}{y}dy<\int_1^{1+x}1dy=x.$$ And we have $$-\log \frac {x}{1+x}=\int^1_{x/(1+x)}\frac {1}{y}dy<\int^1_{x/(1+x)}1 dy=1-\frac {x}{1+x}=\frac {1}{1+x}\implies$$ $$\log \frac {x}{1+x}>\frac {1}{1+x}.$$ For $k>0$ : From the first inequality we have $$\log (1+1/k)<\frac {1}{k}.$$ From the second inequality we have $$\log (1+1/k)= \log \frac {k}{(1+k)}>\frac {1}{1+k}.$$