Proving an inequality between $\frac 1{n+1}$ and $\frac 1n$ and a definite integral For all natural numbers $n$, prove that $$\frac 1{n+1} < \int_n^{n+1} \frac 1t \, dt < \frac 1n$$ 
I have tried working with $\frac 1{t+1} < \frac 1t < \frac 1{t-1}$ but this doesn't help. The result I obtained was $$\log (\frac{k+2}{k+1}) < \int_n^{n+1} \frac 1t \, dt < \log (\frac k{k-1})$$

If $n < t < n+1 \to \dfrac{1}{n+1} < \dfrac{1}{t} < \dfrac{1}{n} \to \dfrac{1}{n+1} = \displaystyle \int_{n}^{n+1}\dfrac{1}{n+1} dt < \displaystyle \int_{n}^{n+1} \dfrac{1}{t}dt<\displaystyle \int_{n}^{n+1}\dfrac{1}{n}dt=\dfrac{1}{n}$
