sum ∑ n = 2 ∞ ( − 1 ) n n 2 + n − 2 I am not able to do the telescoping process in the above series. I converted it into the following partial fraction: ∑ n = 2 ∞ ( − 1 ) n ( n + 2 ) ( n − 1 ) But nothing seems to cancel (as usually happens) using the telescoping method. How can I solve the above series? Is there any other method to do the above problem?

Question

sum             ∑              n        =        2                    ∞                            (        −        1                  )          n                                      n          2                +        n        −        2            I am not able to do the telescoping process in the above series. I converted it into the following partial fraction:      ∑          n      =      2              ∞                  (      −      1              )        n                    (      n      +      2      )      (      n      −      1      )      But nothing seems to cancel (as usually happens) using the telescoping method. How can I solve the above series? Is there any other method to do the above problem?

rae2721 · Accepted Answer

You need to fully expand the partial fraction.                                          (            −            1                          )              n                                            (            n            +            2            )            (            n            −            1            )                                              =                                            (              (              n              +              2              )              −              (              n              −              1              )              )              ×              (              −              1                              )                n                                                    3              ×              (              n              +              2              )              (              n              −              1              )                                                                        =                                            (              n              +              2              )              ×              (              −              1                              )                n                                                    3              ×              (              n              +              2              )              (              n              −              1              )                                      −                                            (              n              −              1              )              ×              (              −              1                              )                n                                                    3              ×              (              n              +              2              )              (              n              −              1              )                                                                        =                                            (              −              1                              )                n                                                    3              ×              (              n              −              1              )                                      −                                            (              −              1                              )                n                                                    3              ×              (              n              +              2              )                                          Now this can be handled using Alternating Harmonic Series.The final result, as Mathematica calculated, is             −      5      +      12      log      ⁡      2        18

Conrad Beltran · Accepted Answer

Writing the nth term of your sum in a form that telescopes can be avoid altogether by converting the problem to a double integral as follows.Noting that      1          n      −      1        =      ∫    0    1        x          n      −      2          d  x    and        1          n      +      2        =      ∫    0    1        y          n      +      1          d  y  ,the sum can be rewritten as                              ∑                      n            =            2                    ∞                                      (            −            1                          )              n                                                          n              2                        +            n            −            2                                              =                  ∑                      n            =            2                    ∞                                      (            −            1                          )              n                                            (            n            −            1            )            (            n            +            2            )                                              (1)                                  =                  ∫          0          1                          ∫          0          1                          ∑                      n            =            2                    ∞                (        −        1                  )          n                          x                      n            −            2                                    y                      n            +            1                                  d        x        d        y                                          =                  ∫          0          1                          ∫          0          1                          y                      x            2                                    ∑                      n            =            2                    ∞                (        −        x        y                  )          n                        d        x        d        y                            (2)                                  =                  ∫          0          1                          ∫          0          1                          y                      x            2                          ⋅                                            x              2                                      y              2                                            1            +            x            y                                  d        x        d        y                                          =                  ∫          0          1                          ∫          0          1                                      y            3                                1            +            x            y                                  d        x        d        y                                          =                  ∫          0          1                          y          2                                      [                          ln        ⁡        (        1        +        x        y        )                                            ]                                0          1                        d        y                                          =                  ∫          0          1                          y          2                ln        ⁡        (        1        +        y        )                d        y                            (3)                                  =                  1          3                ln        ⁡        (        2        )        −                  1          3                          ∫          0          1                                      y            3                                1            +            y                                  d        y                            (4)                                  =                  1          3                ln        ⁡        (        2        )        −                  1          3                          ∫          0          1                          (                      y            2                    −          y          +          1          −                      1                          1              +              y                                )                        d        y                                          =                  1          3                ln        ⁡        (        2        )        −                  1          3                                      [                                          y                3                            3                        −                                          y                2                            2                        +            y            −            ln            ⁡            (            1            +            y            )            ]                    0          1                                                  =                  2          3                ln        ⁡        (        2        )        −                  5          18                .            Explanation1. Interchanging the summation with the double integration.2. Summing the series which is geometric.3. Integration by parts.4. Partial fraction decomposition.

I have the following series: sum _(n=2)^oo((-1)^n)/(n^2+n-2) I am not able to do the telescoping process in the above series. I converted it into the following partial fraction: sum _(n=2)^oo((-1)^n)/((n+2)(n-1))

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