Can You Prove the Inequality Challenge VI for Arctan Sequences?

[anemone]

If $\alpha_n=\arctan n$, prove that $\alpha_{n+1}-\alpha_n<\dfrac{1}{n^2+n}$ for $n=1,\,2,\,\cdots$.

 

[Albert1]

If $\alpha_n=arc\tan n$, prove that $\alpha_{n+1}-\alpha_n<\dfrac{1}{n^2+n}$ for $n=1,\,2,\,\cdots$.

Spoiler

$tan\,\alpha_n=n$, and ,$tan\,\alpha_{n+1}=n+1 $ for $n=1,2,\,\cdots$
$tan(\,\alpha_{n+1}-\,\alpha_n)=\dfrac{1}{n^2+n+1}---(1)$
$\dfrac{1}{n^2+n}---(2)$
$tan\,\dfrac{1}{n^2+n}---(3)$
comare (1)(2)(3) and we prove it
(3)>(2)>(1) for n=1,2,...

 

[anemone]

Spoiler

$tan\,\alpha_n=n$, and ,$tan\,\alpha_{n+1}=n+1 $ for $n=1,2,\,\cdots$
$tan(\,\alpha_{n+1}-\,\alpha_n)=\dfrac{1}{n^2+n+1}---(1)$
$\dfrac{1}{n^2+n}---(2)$
$tan\,\dfrac{1}{n^2+n}---(3)$
comare (1)(2)(3) and we prove it
(3)>(2)>(1) for n=1,2,...

Thanks Albert for participating and your solution! Well done, Albert!:)