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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.

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    $\begingroup$ It should be $<1$, once you divide by $v_n-u_n$, which is fine. $\endgroup$ – carmichael561 Sep 29 '16 at 19:38
  • $\begingroup$ @carmichael561, I edited. Thanks. $\endgroup$ – user231343 Sep 29 '16 at 19:41
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$$\frac{\arctan v-\arctan u}{v-u}=\frac{1}{v-u}\int_{u}^{v}\frac{dz}{1+z^2}\color{red}{<}\frac{1}{v-u}\int_{u}^{v}dz=1\tag{1} $$ since the function $g(z)=\frac{1}{1+z^2}$ attains its absolute maximum, $1$, at a single point ($z=0$).

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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…

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  • $\begingroup$ what about w=0? $\endgroup$ – user231343 Sep 29 '16 at 19:43
  • $\begingroup$ @Edi That's the only case you need to worry about. $\endgroup$ – egreg Sep 29 '16 at 19:44
  • $\begingroup$ Must $\le$ be replaced with $<$ in $\arctan v - \arctan u < v - u$? $\endgroup$ – user231343 Sep 29 '16 at 19:49

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