I want to show that the series ∑ n ≥ 1 1 ( n + 1 ) a + 1 ∑ k = 0 n b k ( ( n − k ) ! n ! ) a

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I want to show that the series      ∑          n      ≥      1            1          (      n      +      1              )                  a          +          1                          ∑          k      =      0        n        b    k              (                        (          n          −          k          )          !                          n          !                    )        a

tilsjaskak6 · Accepted Answer

It is enough to show that the sum for   n  ≥  0 converges. Changing sums and manipulating I get:      ∑          k      ≥      0                  b      k              k      !            ∑          n      ≥      k            1          (      n      +      1              )                  a          +          1                            (                                            n                                                          k                                      )              ≤      ∑          k      ≥      0                  b      k              k      !            ∑          n      ≥      k            1          (      n      +      1              )                  a          +          1                      ≤      ∑          k      ≥      0                  b      k              k      !            ∑          n      ≥      0            1          (      n      +      1              )                  a          +          1                      =      ∑          k      ≥      0                  b      k              k      !        C  =  C      e          b        =      ∑          k      ≥      0                  b      k              k      !        C  =  C      e          b      where C is a constant   &amp;lt;  ∞ because   a  +  1  &amp;gt;  1

Addison Trujillo · Accepted Answer

For convenience, we consider the sum starting at n=0. Then                              ∑                      n            =            0                    ∞                          1                      (            n            +            1                          )                              a                +                1                                                              ∑                      k            =            0                    n                          b          k                                      (                                          (                n                −                k                )                !                                            n                !                                      )                    a                                    =                              ∑                      k            =            0                    ∞                                      b            k                                (            k            !                          )              a                                                ∑                      n            =            k                    ∞                          1                      (            n            +            1                          )                              a                +                1                                                              1                                                                                          (                                                                    n                  k                                                                      )                                                                                      a                                                                        ≤                              ∑                      k            =            0                    ∞                                      b            k                                (            k            !                          )              a                                                ∑                      n            =            k                    ∞                          1                      (            n            +            1                          )                              a                +                1                                                                                      ≤                    ζ        (        a        +        1        )                  ∑                      k            =            0                    ∞                                      b            k                                (            k            !                          )              a                                      .

I want to show that the series <munder> ∑ <mrow class="MJX-TeXAtom-ORD"

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New Questions in Calculus 2