# Prove that $1/n+1/(n+1)+\dots+(1/2n)>2/3$

Prove that $$\displaystyle\frac{1}{n} + \frac{1}{n+1} + \dots + \frac{1}{2n}>\frac{2}{3}$$

I tried to use mathematical induction, but I'm not able to prove that: $$\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}>\frac{2}{3}.$$

My method was:

Assumption: $$\frac{1}{n} + \frac{1}{n+1} + \dots + \frac{1}{2n}>\frac{2}{3}$$

$$\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}>\frac{2}{3} + \frac{1}{2n+1} + \frac{1}{2n+2}$$

$$\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}>\frac{2}{3} + \frac{1}{2n+1} + \frac{1}{2n+2}-\frac{1}{n}$$

Now in order to prove the thesis I have to prove that $$\frac{2}{3} + \frac{1}{2n+1} + \frac{1}{2n+2}-\frac{1}{n} > \frac{2}{3}$$

But it's a contradiction. Did I make a mistake somewhere? How can I solve this problem? I'd appreciate your help.

• Hint: try to prove a stronger inequality by omitting the first term of the sum. As if by miracle, induction will work then! (note: this inequality will only hold for large $n$) Oct 28, 2018 at 20:53
• Actually, it should work for every natural n. I will try your idea, thank you a lot!
– user609637
Oct 28, 2018 at 20:57
• Also, to explicitly comment on your approach: there isn't a mistake there. It's just that sometimes naively using induction simply won't work. In such cases you either need to somehow strengthen the induction hypothesis (like I did), or take a completely different approach. Oct 28, 2018 at 21:07

HINT

As suggested by Wojowu in the comment, sometimes induction works for a stronger hypothesis, in that case let try with

$$\displaystyle\frac{1}{n} + \frac{1}{n+1} + \dots + \frac{1}{2n}>\frac{2}{3}+\frac1{4n}>\frac23$$

Refer also to the related

There is also a "non-inductive" way to show the inequality.

You may transform the sum into a Riemann-sum and then estimate this sum by an integral: $$1/n+1/(n+1)+\dots+(1/2n) = \frac{1}{n}\left(\frac{1}{1+\frac{0}{n}} + \frac{1}{1+\frac{1}{n}} + \cdots + \frac{1}{1+\frac{n}{n}} \right) =\boxed{\sum_{k=0}^n \frac{1}{1+\frac{k}{n}} \cdot \frac{1}{n}}$$ Now, note that $$f(x) = \frac{1}{1+x}$$ is strictly decreasing $$[0,1]$$. So, you have

• $$\frac{1}{1+\frac{k}{n}} \cdot \frac{1}{n} = \int_{\frac{k}{n}}^{\frac{k+1}{n}} \frac{1}{1+\frac{k}{n}} \; dx > \int_{\frac{k}{n}}^{\frac{k+1}{n}} \frac{1}{1+x}\; dx$$

This gives $$\boxed{\sum_{k=0}^n \frac{1}{1+\frac{k}{n}} \cdot \frac{1}{n} > \int_0^1 \frac{1}{1+x}\; dx = \ln 2 > \frac{2}{3}}$$