Prove that $\frac 12\leq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}...+\frac{1}{n+n}$

Question

How do I prove $$\frac 12\leq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}...+\frac{1}{n+n}$$

without using induction? Note that clearly $n\neq 0$

Thanks for any help!!

Answer 1

This inequality does not hold since for any $n\in \mathbb N$ you have $$\frac{1}{2}\leq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+n}.$$ To see this, note first that any term above is greater than $\frac{1}{n+n}$, so that your sum is greater than $n\cdot \frac{1}{2n}=\frac{1}{2}.$

Answer 2

The reverse inequality is true for $n≥1$.

If $1≤k≤n$ then $\frac{1}{n+k}≥\frac{1}{n+n}$. Therfore, you get :

$$\frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}...+\frac{1}{n+n} ≥ \frac{1}{n+n}+\frac {1}{n+n}+\frac{1}{n+n}...+\frac{1}{n+n} = n \cdot \frac{1}{n+n} = \frac{1}{2}$$