Consider the infinite sum: ∑ k = 0 ∞ ( − 1 ) k x 3 k + 2 ( 3 k + 2 ) ! .

Question

Consider the infinite sum:      ∑          k      =      0              ∞                  (      −      1              )                  k                            x                  3          k          +          2                            (      3      k      +      2      )      !        .

Syllingbs · Accepted Answer

Define the following three sums:      s    1    (  x  )  =      ∑          k      =      0              ∞                  (      −      1              )        k                    x        k                    k      !            s    2    (  x  )  =      ∑          k      =      0              ∞                  (      −      1              )        k            (      ω      x              )        k                    k      !            s    3    (  x  )  =      ∑          k      =      0              ∞                  (      −      1              )        k            (              ω        2            x              )        k                    k      !      If we add these three equations up, the       x          3      n      +      1       and       x          3      n      +      2       terms vanish because of the identity   1  +  ω  +      ω    2    =  0, and we’re left with the following:      s    1    (  x  )  +      s    2    (  x  )  +      s    3    (  x  )  =  3      ∑          k      =      0              ∞                  (      −      1              )        k                    x                  3          k                            (      3      k      )      !      This is close to what we want, except we’ve now “selected” only the   3  n  +  0 terms instead of the   3  n  +  2 terms. We can fix this by adding scaled versions of our sums:      s    1    (  x  )  +  ω      s    2    (  x  )  +      ω    2        s    3    (  x  )  =  3      ∑          k      =      0              ∞                  (      −      1              )                  k                            x                  3          k          +          2                            (      3      k      +      2      )      !      Notice that again we’ve used the fact that   1  +  ω  +      ω    2    =  0, however by multiplying each sequence by some power of   ω, we make that product only take place in the       x          3      n       and       x          3      n      +      1       terms.This tells us that:      ∑          k      =      0              ∞                  (      −      1              )        k                    x                  3          k          +          2                            (      3      k      +      2      )      !        =                           s        1            (      x      )      +      ω              s        2            (      x      )      +              ω        2                    s        3            (      x      )        3  We finish this by noticing that       s    1    (  x  )  =      e          −      x      ,       s    2    (  x  )  =      e          −      ω      x      , and       s    3    =      e          −              ω        2            x      . So our final answer is:                    e                  −          x                    +      ω              e                  −          ω          x                    +              ω        2                    e                  −                      ω            2                    x                      3

Consider the infinite sum: <munderover> ∑ <mrow class="MJX-TeXAtom-ORD"

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