Prove $\frac{1}{n+1}\le \log\left(1+\frac{1}{n}\right)\le \frac{1}{n} ,\forall n\ge1$ if $\log x= \int_{1}^{x}\frac{dt}{t},x>0$ I know it is very simple but something is going amiss.
We can see here that $\log(1+\dfrac{1}{n})=\displaystyle \int_{n}^{n+1}\dfrac{dt}{t}$
if $t \in[n,n+1]$, $\dfrac{1}{n+1}\le\dfrac{1}{t}\le\dfrac{1}{n}$
I want to show that $\dfrac{1}{n+1}\le\displaystyle \int_{n}^{n+1}\dfrac{dt}{t}\le\dfrac{1}{n}$.
Can I straight infer that? If so what is the logic behind it? I was thinking of using $\displaystyle \Sigma_{i=1}^{n}$ and sandwiching $\displaystyle \int_{n}^{n+1}\dfrac{dt}{t}$ between. But confused !

$$n\le t\le n+1$$
$$\frac {1}{n+1} \le \frac {1}{t}\le \frac {1}{n}$$
$$ \int _n ^{n+1}\frac {dt}{n+1} \le \int _n ^{n+1}\frac {dt}{t}\le \int _n ^{n+1}\frac {dt}{n}$$
$$\dfrac{1}{n+1}\le\displaystyle \int_{n}^{n+1}\dfrac{dt}{t}\le\dfrac{1}{n}$$
