\frac{1}{n+1}+ \frac{1}{n+2}+ \cdot\cdot\cdot + \frac{1}{2n}\le \int_{2n}^n\frac{dt}{t}\le \frac{1}{n}+ \cdot\cdot\cdot + \frac{1}{2n-1}
we must show this not deduce it
\frac{1}{n+1}+ \frac{1}{n+2}+ \cdot\cdot\cdot + \frac{1}{2n}\le \int_{2n}^n\frac{dt}{t}\le \frac{1}{n}+ \cdot\cdot\cdot + \frac{1}{2n-1}
its like this
not two integrals its one integral
show that 1/n+1 +1/n+2 +...1/2n≤from 2n to n∫dt/t≤1/n+...1/2n-1
consider the sequene (Vn) defined by:
Vn= 1/n+1 +...+1/2n=from 2n to p=n+1∑1/p
deduce from above that ln2-1/2n≤ Vn≤ln2
i didnt find any thing