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 $$\log (\frac{k+2}{k+1}) < \int_n^{n+1} \frac 1t \, dt < \log (\frac k{k-1})$$

For $t \in (n, n+1)$, $$\frac{1}{n +1} < \frac{1}{t} < \frac{1}{n}.$$
Integrate from $n$ to $n + 1$ to get the desired result. (well done improving your question)
