Prove that $\frac{1}{k+1} \le \ln(1+\frac{1}{k}) \le \frac{1}{k}$ I am trying to prove that $\frac{1}{k+1} \le\ln(1+\frac{1}{k}) \le \frac{1}{k}$. This showed up as a known fact in a proof for $(1+\frac{1}{k})^k = e$ as $k$ goes to infinity, so I would like to prove it without using that fact.
I tried to prove the first part of the inequality by using another known inequality (one which I am comfortable proving):
$1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k} >\ln(1+k)$
Let $a$ denote $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k-2}$
$a+\frac{1}{k-1}+\frac{1}{k} >\ln(1+k)$
Scaling this down 1 term:
$a + \frac{1}{k-1} >\ln(k)$
Subtracting the second inequality from the first, we get:
$\frac{1}{k} >\ln(1+\frac{1}{k})$
However, this seems to contradict the statement I'm trying to prove. Could anyone help?
 A: Hint:
$$\ln\left(1 + \frac1k \right) = \int_k^{k+1}\frac1t dt$$
A: The Mean Value Theorem says that for some $0\lt\xi\lt1$
$$
\log(k+1)-\log(k)=\frac1{k+\xi}
$$
Therefore
$$
\frac1{k+1}\lt\log\left(1+\frac1k\right)\lt\frac1k
$$
A: Your mistake is that subtracting like inequalities doesn't tell you anything useful: for example


*

*$2<4$ and $1<2$, and we do have $2-1 < 4-2$

*$2<4$ and $1<3$, but we have $2-1 = 4-3$

*$2<3$ and $1<3$, but we have $2-1 > 0$


Knowing $a<b$ and $c<d$ doesn't let us conclude anything about how $a-c$ and $b-d$ compare.
Subtracting opposite ones is useful, though:

Theorem: If $a<b$ and $c>d$, then $a-c < b-d$.

However, in my opinion, it's easier to remember just that adding like inequalities is fine, and convert $c>d$ to $-c<-d$ when desired.
A: A variant of
Open Ball's answer:
$\ln\left(1 + \frac1 k \right) 
= \int_1^{1+\frac1{k}}\frac{dt} t
= \int_0^{\frac1{k}}\frac{dt} {1+t} 
$
so
$\ln\left(1 + \frac1 k \right) 
< \int_0^{\frac1{k}}\frac{dt} {1} 
=\frac1{k}
$
and
$\ln\left(1 + \frac1 k \right) 
> \int_0^{\frac1{k}}\frac{dt} {1+1/k} 
=\frac1{k}\frac1 {1+1/k} 
=\frac1{k+1}
$.
