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

Question

I wanted to prove this by mathematical induction. I have proved that it is less than two and that the sum is greater than 0 but I am stuck on how to prove it to be greater than 1. Please help!!!

I would like a conclusive proof for this. Thank you so much!

I do believe there was an answer to a question similar to this, but the answers provided there did not use mathematical induction to provide the answer.

Answer 1

This is a matter of comparison testing.

First render $n+2>n+1>0, n+3>n+1>0, \text{ etc}$. Thereby

$\dfrac{1}{n+1}+\dfrac{1}{n+2}+\dfrac{1}{n+3}+...+\dfrac{1}{3n+2}<\dfrac{1}{n+1}+\dfrac{1}{n+1}+\dfrac{1}{n+1}+...+\dfrac{1}{n+1}$

where the right side has $2n+2=2(n+1)$ identical terms. So

$\color{blue}{\dfrac{1}{n+1}+\dfrac{1}{n+2}+\dfrac{1}{n+3}+...+\dfrac{1}{3n+2}<\dfrac{2(n+1)}{n+1}=2}$

The lower bound is trickier. Here we start with the following for $a>b>0$:

$\dfrac{1}{a-b}+\dfrac{1}{a+b}=\dfrac{2a}{a^2-b^2}>\dfrac{2a}{a^2}=\dfrac{2}{a}$

Apply this with $a=2n+1$ and, successively, $b=1, b=2, n=3, ..., b=n$. Thereby

$\dfrac{1}{2n}+\dfrac{1}{2n+2}>\dfrac{2}{2n+1}$

$\dfrac{1}{2n-1}+\dfrac{1}{2n+3}>\dfrac{2}{2n+1}$

$\dfrac{1}{2n-2}+\dfrac{1}{2n+4}>\dfrac{2}{2n+1}$

...

$\dfrac{1}{n+1}+\dfrac{1}{3n+2}>\dfrac{2}{2n+1}$

Add these up:

$\color{blue}{\dfrac{1}{n+1}+\dfrac{1}{n+2}+\dfrac{1}{n+3}+...+\dfrac{1}{3n+2}>2(\dfrac{1}{2n+1})+2(\dfrac{1}{2n+1})+2(\dfrac{1}{2n+1})+...+\dfrac{1}{2n+1}=n(2(\dfrac{1}{2n+1}))+\dfrac{1}{2n+1}=\dfrac{2n+1}{2n+1}=1}$.

Answer 2

The first inequality is equivalent to $$\frac{2(n+1)}{\sum_{k=1}^{2(n+1)} \frac{1}{n+k}} < \frac{4(n+1)^2}{2(n+1)}$$ Now see that $\sum_{k=1}^{2(n+1)} n+k = 4(n+1)^2$ so the lefthand side is the harmonic mean and the righthand side the arithmetic mean. Since the individual terms in the sum are not equal the inequality is proven by AM-HM.