Determine lim n → ∞ arctan ⁡ ( n + 1 ) − arctan ⁡ ( n )

Question

Determine       lim          n      →      ∞        arctan  ⁡  (            n      +      1        )  −  arctan  ⁡  (            n        )

Julien Carrillo · Accepted Answer

Note,  arctan  ⁡            n      +      1        −  arctan  ⁡            n        =  arctan  ⁡                    n        +        1            −              n                    1      +              n        +        1            ⋅              n              =  arctan  ⁡      1          (      1      +                        n          2                +        n            )      (              n        +        1            +              n            )      Thus,      lim          n      →      ∞        arctan  ⁡  (            n      +      1        )  −  arctan  ⁡  (            n        )  =      lim          n      →      ∞        arctan  ⁡      1          (      1      +                        n          2                +        n            )      (              n        +        1            +              n            )        =  arctan  ⁡  (  0  )  =  0

Nicholas Cruz · Accepted Answer

The arctan function is the inverse function of

\tan : (- \frac{π}{2}, \frac{π}{2}) \to R

as this function is monotonically increasing, we have

lim_{x \to \frac{π}{2}} \tan x = + \infty ⟺ lim_{x \to + \infty} \arctan x = \frac{π}{2}

therefore

lim_{n \to \infty} \arctan (\sqrt{n + 1}) = \frac{π}{2}

lim_{n \to \infty} \arctan (\sqrt{n}) = \frac{π}{2}

so the difference is 0.

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