Finding lim n → ∞ tan ⁡ ( tan − 1 ⁡ 2 1 2 + tan − 1 ⁡ 2 2 2 + . . . + tan − 1 ⁡ 2 n 2 ) Using the identity I rewrote tan − 1 ⁡ 2 1 2 + tan − 1 ⁡ 2 2 2 + . . . + tan − 1 ⁡ 2 n 2 as; tan − 1 ⁡ ( 2 ) − tan − 1 ⁡ ( 0 ) + tan − 1 ⁡ ( 3 ) − tan − 1 ⁡ ( 1 ) + tan − 1 ⁡ ( 4 ) − tan − 1 ⁡ ( 2 ) + . . . + tan − 1 ⁡ ( n + 1 ) − tan − 1 ⁡ ( n − 1 ) Which simplified to − tan − 1 ⁡ ( 0 ) − tan − 1 ⁡ ( 1 ) + tan − 1 ⁡ ( n + 1 ) Which simplified to tan − 1 ⁡ ( n n + 2 ) Therefore the original limit is now lim n → ∞ n n + 2 = 1However, the answers say the limit approaches −1 and using my calculator I found that the tan-inverse chain approaches 3 π 4 which supports the answer given in the book. So where did I go wrong?

Question

Finding       lim          n      →      ∞        tan  ⁡            (              tan                  −          1                    ⁡                        2                      1            2                              +              tan                  −          1                    ⁡                        2                      2            2                              +      .      .      .      +              tan                  −          1                    ⁡                        2                      n            2                              )      Using the identity I rewrote       tan          −      1        ⁡            2              1        2              +      tan          −      1        ⁡            2              2        2              +  .  .  .  +      tan          −      1        ⁡            2              n        2             as;      tan          −      1        ⁡      (    2    )    −      tan          −      1        ⁡      (    0    )    +      tan          −      1        ⁡      (    3    )    −      tan          −      1        ⁡      (    1    )    +      tan          −      1        ⁡      (    4    )    −      tan          −      1        ⁡      (    2    )    +  .  .  .  +      tan          −      1        ⁡      (    n    +    1    )    −      tan          −      1        ⁡      (    n    −    1    )  Which simplified to   −      tan          −      1        ⁡      (    0    )    −      tan          −      1        ⁡      (    1    )    +      tan          −      1        ⁡      (    n    +    1    )  Which simplified to       tan          −      1        ⁡            (              n                  n          +          2                    )      Therefore the original limit is now       lim          n      →      ∞            n          n      +      2        =  1However, the answers say the limit approaches −1 and using my calculator I found that the tan-inverse chain approaches             3      π        4   which supports the answer given in the book. So where did I go wrong?

rcampas4i · Accepted Answer

Wait, what happened to       tan          −      1        ⁡  n? Your sum telescopes, but the lag is by 2 terms, not 1. In particular,      ∑          k      =      1        n        tan          −      1        ⁡      2          n      2        =      ∑          k      =      1        n        tan          −      1        ⁡  (  n  +  1  )  −      tan          −      1        ⁡  (  n  −  1  )  =      ∑          k      =      2              n      +      1            tan          −      1        ⁡  n  −      ∑          k      =      0              n      −      1            tan          −      1        ⁡  n    =      tan          −      1        ⁡  n  +      tan          −      1        ⁡  (  n  +  1  )  −      tan          −      1        ⁡  0  −      tan          −      1        ⁡  1.

Finding lim_(n -> oo) tan(tan^(-1) 2/(1^2)+tan^(-1) 2/(2^2)+...+tan^(-1) 2/(n^2))

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