I recently came across the result that ∑ n = 2 ∞ n 4 − n 3 + n + 1 n 6 − 1 = 1 2 I am wondering how one could proof this, generally how one could evaluate a sum over rational functions.If I plug the sum into Wolfram Alpha it gives 3 k 4 − k − 2 6 k ( k + 1 ) ( k 2 + k + 1 ) as the k-th partial sum. Taking the limit as n → ∞, this would in fact proof the upper equality.Sadly, I could not wrap my head around how to get to Wolfram Alphas partial sum result. If anyone has an idea let me know. Any tips are appreciated.

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I recently came across the result that      ∑          n      =      2        ∞                      n        4            −              n        3            +      n      +      1                      n        6            −      1        =      1    2  I am wondering how one could proof this, generally how one could evaluate a sum over rational functions.If I plug the sum into Wolfram Alpha it gives            3              k        4            −      k      −      2              6      k      (      k      +      1      )      (              k        2            +      k      +      1      )      as the   k-th partial sum. Taking the limit as   n  →  ∞, this would in fact proof the upper equality.Sadly, I could not wrap my head around how to get to Wolfram Alphas partial sum result. If anyone has an idea let me know. Any tips are appreciated.

AlokMoopisppf · Accepted Answer

In a word: the terms telescope in a nice way.Using partial fractions, we have                    n        4            −              n        3            +      n      +      1                      n        6            −      1        =      1    3        (                  n        −        2                              n          2                −        n        +        1              −                  n        −        1                              n          2                +        n        +        1              +          1              n        −        1              −          1              n        +        1              )  We need to be a bit careful because the harmonic series diverges, so we should only group terms with opposite signs together. The last two terms form a well-known telescoping series:      1    3        ∑          n      =      2              m            (          1              n        −        1              −          1              n        +        1              )    =      1    3        (    1    +          1      2        −          1      m        −          1              m        +        1              )    ⟶      1    2  The first two terms telescope as well, with the second term eating the one before it:      1    3        ∑          n      =      2              m            (                  n        −        2                              n          2                −        n        +        1              −                  n        −        1                              n          2                +        n        +        1              )    =  −      1    3    ⋅            m      −      1                      m        2            +      m      +      1        ⟶  0This sort of thing won&#039;t work in general but I won&#039;t look a gift horse in the mouth.

I recently came across the result that <munderover> ∑ <mrow class="M

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