I'm trying to show that u n = ∑ k = 1 n ( − 1 ) k k ∼ n → ∞ ( − 1 ) n n 2 when n → ∞

Question

I&#039;m trying to show that      u          n        =      ∑          k      =      1        n    (  −  1      )    k        k        ∼          n      →      ∞        (  −  1      )    n              n        2  when   n  →  ∞

scoseBexgofvc · Accepted Answer

We have estimations      u          2      n        =      ∑          k      =      1              2      n        (  −  1      )    k        k    =      ∑          k      =      1              n        (      2    k    −      2    k    −    1    )  =      ∑          k      =      1              n            1                  2        k            +              2        k        −        1              ≤      ∑          k      =      1              n            1          2              2        k        −        1              ≤      ∫          1        n              d      x              2              2        (        x        +        1        )        −        1              =                    2        n        +        1            −              3              2  and      u          2      n        =      ∑          k      =      1              2      n        (  −  1      )    k        k    =      ∑          k      =      1              n        (      2    k    −      2    k    −    1    )  =      ∑          k      =      1              n            1                  2        k            +              2        k        −        1              ≥      ∑          k      =      1              n            1          2              2        k              ≥      ∫          1        n              d      x              2              2        x              =                    2        n            −              2              2  hence       u          2      n        ∼      1    2        2    n   while   n  →  ∞

Jeramiah Campos · Accepted Answer

Norbert&#039;s answer is the simplest with the fewest prerequisites. That being the case, here is a nuke for an ant hill using the Euler-Maclaurin Sum Formula.Note that                              ∑                      k            =            1                                2            n                          (        −        1                  )          k                          k                                    =        2                  ∑                      k            =            1                                n                                    2          k                        −                          ∑                      k            =            1                                2            n                                    k                                    (1)                                  =                  8                          ∑                      k            =            1                                n                                    k                        −                          ∑                      k            =            1                                2            n                                    k                    The Euler-Maclaurin Sum Formula says that                    (2)                              ∑                      k            =            1                    n                          k                =                  2          3                          n                      3                          /                        2                          +                  1          2                          n                      1                          /                        2                          +        C        +                  1          24                          n                      −            1                          /                        2                          +        O                  (                      n                          −              3                              /                            2                                )                    For       R    e    (  z  )  &amp;gt;  −  1                    (3)                    ζ        (        z        )        =                  lim                      n            →            ∞                                    ∑                      k            =            1                    n                          k                      −            z                                  −                          (                      1                          1              −              z                                            n                          1              −              z                                +                      1            2                                n                          −              z                                )                    Applying (3) to (2) yields   C  =  ζ  (  −      1    2    )  =  −      1          4      π        ζ  (      3    2    )              =      ˙        −  0.207886224977354566Applying (2) to (1) yields                              ∑                      k            =            1                                2            n                          (        −        1                  )          k                          k                                    =                  8                          ∑                      k            =            1                                n                                    k                        −                          ∑                      k            =            1                                2            n                                    k                                                  =                  8                          (                      2            3                                n                          3                              /                            2                                +                      1            2                                n                          1                              /                            2                                +          C          +                      1            24                                n                          −              1                              /                            2                                )                                                  −                  (                      8                                2            3                                n                          3                              /                            2                                +                      2                                                    1            2                                n                          1                              /                            2                                +          C          +                      1                          2                                            1            24                                n                          −              1                              /                            2                                )                +        O                  (                      n                          −              3                              /                            2                                )                                    (4)                                  =                  1                      2                                    n                      1                          /                        2                          +        (                  8                −        1        )        C        +                  3                      2                                    1          24                          n                      −            1                          /                        2                          +        O                  (                      n                          −              3                              /                            2                                )                    Thus, for even n, we get                    (5)                              ∑                      k            =            1                                n                          (        −        1                  )          k                          k                =                  1          2                          n                      1                          /                        2                          +                  1          8                          n                      −            1                          /                        2                          +                              7            C                                              8                        +            1                          +        O                  (                      n                          −              3                              /                            2                                )                    For odd n, we get                              ∑                      k            =            1                                n                          (        −        1                  )          k                          k                                    =                  ∑                      k            =            1                                n            −            1                          (        −        1                  )          k                          k                        −                          n                                                  =                              1            2                    (          n          −          1                      )                          1                              /                            2                                      +                              1            8                    (          n          −          1                      )                          −              1                              /                            2                                      −                  n                      1                          /                        2                          +                              7            C                                              8                        +            1                          +        O                  (                      n                          −              3                              /                            2                                )                                                  =                              1            2                                n                          1                              /                            2                                −                      1            4                                n                          −              1                              /                            2                                      +                              1            8                                n                          −              1                              /                            2                                      −                  n                      1                          /                        2                          +                              7            C                                              8                        +            1                          +        O                  (                      n                          −              3                              /                            2                                )                                    (6)                                  =        −                  1          2                          n                      1                          /                        2                          −                  1          8                          n                      −            1                          /                        2                          +                              7            C                                              8                        +            1                          +        O                  (                      n                          −              3                              /                            2                                )                    using the Binomial Theorem to expand the powers of   (  n  −  1  )Combining (5) and (6), we get                    (7)                              ∑                      k            =            1                                n                          (        −        1                  )          k                          k                =        (        −        1                  )          n                          (                      1            2                                n                          1                              /                            2                                +                      1            8                                n                          −              1                              /                            2                                )                −                  7                                    8                        +            1                                                ζ            (                          3              2                        )                                4            π                          +        O                  (                      n                          −              3                              /                            2                                )                    We&#039;ve used more powerful machinery, but we&#039;ve also gotten a more precise answer.

I'm trying to show that u <mrow class="MJX-TeXAtom-ORD"> n </mrow> </msub>

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New Questions in Calculus 2