# Prove that $\arctan v - \arctan u < v - u$ if $u<v$.

The following is the last exercise of chapter 5 of the Advanced Calculus by Fitzpatrick:

Prove that $\arctan v - \arctan u < v - u$ if $u<v$.

Suppose two sequences $\{u_n\}$ and $\{v_n\}$. If the $$\dfrac{\arctan v_n - \arctan u_n}{v_n - u_n} <1 \quad \text{ for all indices } n,$$ But I don't know how to proceed.

• It should be $<1$, once you divide by $v_n-u_n$, which is fine. – carmichael561 Sep 29 '16 at 19:38
• @carmichael561, I edited. Thanks. – user231343 Sep 29 '16 at 19:41

$$\frac{\arctan v-\arctan u}{v-u}=\frac{1}{v-u}\int_{u}^{v}\frac{dz}{1+z^2}\color{red}{<}\frac{1}{v-u}\int_{u}^{v}dz=1\tag{1}$$ since the function $g(z)=\frac{1}{1+z^2}$ attains its absolute maximum, $1$, at a single point ($z=0$).
Do the math more carefully: saying $\arctan v-\arctan u<v-u$ is the same as saying $$\frac{\arctan v-\arctan u}{v-u}<1$$ The mean value theorem says $$\frac{\arctan v-\arctan u}{v-u}=\frac{1}{1+w^2}$$ for some $w\in(u,v)$, so…
• Must $\le$ be replaced with $<$ in $\arctan v - \arctan u < v - u$? – user231343 Sep 29 '16 at 19:49