Given the series ∑ k = 1 ∞ 1 k ( k 2 + 1 )

Question

Given the series      ∑          k      =      1              ∞            1                  k                    (                  k          2                +        1        )

Tucker House · Accepted Answer

Using the binomial theorem to expand                                                        1                              k                (                                  k                  2                                +                1                )                                                                        =                  k                      −            3                          /                        2                                                (            1            +                          1                              k                2                                      )                                −            1                          /                        2                                              (1)                                  ∼                  k                      −            3                          /                        2                          −                  1          2                          k                      −            7                          /                        2                          +                  3          8                          k                      −            11                          /                        2                          −                  5          16                          k                      −            15                          /                        2                          +                  35          128                          k                      −            19                          /                        2                          −                  63          256                          k                      −            23                          /                        2                              with an error of approximately       1    4        k          −      27              /            2      Applying the Euler Maclaurin Sum Formula to (1) yields                              ∑                      k            =            1                    n                                                    1                              k                (                                  k                  2                                +                1                )                                                                        ∼        C        −        2                  n                      −            1                          /                        2                          +                  1          2                          n                      −            3                          /                        2                          +                  3          40                          n                      −            5                          /                        2                          −                  1          4                          n                      −            7                          /                        2                          +                  31          384                          n                      −            9                          /                        2                          +                  3          16                          n                      −            11                          /                        2                                                            −                  12959          66560                          n                      −            13                          /                        2                          −                  5          32                          n                      −            15                          /                        2                          +                  214861          557056                          n                      −            17                          /                        2                          +                  35          256                          n                      −            19                          /                        2                                              (2)                                  −                  24553421          27525120                          n                      −            21                          /                        2                          −                  63          512                          n                      −            23                          /                        2                              with an error of approximately   3      n          −      25              /            2      . Since the series is convergent, C represents the sum.Computing the partial sum to   n  =  100 and using (2) to compute C, we get  C                    =                    .              2.264062399141221028592305Just as a check, with   n  =  1000 (2) gives  C                    =                    .              2.264062399141221028592304976971183194

Given the series <munderover> ∑ <mrow class="MJX-TeXAtom-ORD"> k

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Answer & Explanation

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