Inequality Challenge: Prove $\sum_{1}^{n}$

[Albert1]

$n\in N,n\geq 2$

prove:

$ \sum_{1}^{n}(\dfrac{1}{2n-1}-\dfrac{1}{2n})>\dfrac {2n}{3n+1}$

 

[anemone]

$n\in N,n\geq 2$

prove:

$ \sum_{1}^{n}(\dfrac{1}{2n-1}-\dfrac{1}{2n})>\dfrac {2n}{3n+1}$

My solution:

Spoiler

Note that

\(\displaystyle \begin{align*}1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\cdots+\frac{1}{2n-1}-\frac{1}{2n}&=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}\\&=\left(1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}\right)\\&=\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}\right)\\&=\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2n}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)\\&=\small\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)\\&=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\end{align*}\)

Therefore

\(\displaystyle \begin{align*}\sum_{k=1}^{n}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n}\right)&=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\\&\ge \frac{(\overbrace{1+1+\cdots+1}^{\text{n times}})^2}{n+1+n+2+\cdots+2n}\,\,\text{(by the extended Cauchy-Schwarz inequality)}\\&=\frac{n^2}{\frac{n}{2}\left(n+1+2n\right)}\\&=\frac{2n}{3n+1}\,\,\,\,\text{(Q.E.D.)}\end{align*}\)

 

[Albert1]

My solution:

Spoiler

Note that

\(\displaystyle \begin{align*}1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\cdots+\frac{1}{2n-1}-\frac{1}{2n}&=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}\\&=\left(1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}\right)\\&=\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}\right)\\&=\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2n}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)\\&=\small\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)\\&=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\end{align*}\)

Therefore

\(\displaystyle \begin{align*}\sum_{k=1}^{n}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n}\right)&=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\\&\ge \frac{(\overbrace{1+1+\cdots+1}^{\text{n times}})^2}{n+1+n+2+\cdots+2n}\,\,\text{(by the extended Cauchy-Schwarz inequality)}\\&=\frac{n^2}{\frac{n}{2}\left(n+1+2n\right)}\\&=\frac{2n}{3n+1}\,\,\,\,\text{(Q.E.D.)}\end{align*}\)

very good your answer is correct!

 

[Greg]

My solution:

Spoiler

We wish to prove

$$\sum_{i=1}^n\left(\dfrac{1}{2i-1}-\dfrac{1}{2i}\right)\gt\dfrac{2n}{3n+1},\quad n\ge2,\quad n\in\mathbb{N}$$

I claim that

$$\sum_{i=1}^n\dfrac{2}{9i^2-3i-2}=\dfrac{2n}{3n+1}$$

Proof:

$$\dfrac{2}{9(1)^2-3(1)-2}=\dfrac{2(1)}{3(1)+1}=\dfrac12$$
$$\dfrac{2n}{3n+1}-\dfrac{2(n-1)}{3(n-1)+1}=\dfrac{2}{9n^2-3n-2}$$
$$\Rightarrow\dfrac{2n}{3n+1}=\dfrac{2(n-1)}{3(n-1)+1}+\dfrac{2}{9n^2-3n-2}$$

as required.

$$\sum_{i=1}^n\left(\dfrac{1}{4i^2-2i}-\dfrac{2}{9i^2-3i-2}\right)>0\quad\forall\, n>1$$
$$\text{Q.E.D.}$$