I ran into this limit while evaluating an integral. lim N → ∞ ( 2 N + 1 − ∑ n = 1 N 1 n )

Question

I ran into this limit while evaluating an integral.      lim          N      →      ∞            (    2          N      +      1            −              ∑              n        =        1            N              1              n              )

jugf5 · Accepted Answer

First of all we see that eta-function is defined (the series converges) for   s  &amp;gt;  0 (Dirichlet convergence test). Next, we use the relation between eta-function and zeta-function for   s  &amp;gt;  1  η  (  s  )  =      ∑          n      =      1        ∞              (      −      1              )                  n          −          1                            n      s        =      ∑          n      =      1        ∞        1          n      s        −  2      ∑          n      =      1        ∞        1          (      2      n              )        s              =            (        1  −      2          1      −      s                  )        ζ  (  s  )Using this relation we can define zeta-function for   s  ∈  (  0  ;  1  ) as   ζ  (  s  )  =            η      (      s      )              1      −              2                  1          −          s                     We can also present eta-function in the form      lim          N      →      ∞            ∑          n      =      1              2      N                  (      −      1              )                  n          −          1                            n      s        =      lim          N      →      ∞                  (            ∑          n      =      1              2      N            1          n      s        −  2      ∑          n      =      1        N        1          (      2      n              )        s                        )        =      lim          N      →      ∞                  (            ∑          n      =      1              N                  1      −              2                  1          −          s                            n      s        +      ∑          n      =      N      +      1              2      N            1          n      s                  )      The second term can be presented as the Riemann sum at   N  →  ∞      I    2    (  N  )  =      ∑          n      =      N      +      1              2      N            1          n      s        =      N          N      S            1    N        ∑          n      =      N      +      1              2      N            1                            (                            n        N                                          )                          s              →      N          1      −      s            ∫    1    2              d      x              x      s        =      N          1      −      s                          2                  1          −          s                    −      1              1      −      s        η  (  s  )  =      lim          N      →      ∞                  (            ∑          n      =      1              N                  1      −              2                  1          −          s                            n      s        −      N          1      −      s                  1      −              2                  1          −          s                            1      −      s                  )        ζ  (  s  )  =      lim          N      →      ∞                  (            ∑          n      =      1              N            1          n      s        −            N              1        −        s                    1      −      s                  )          ;        s  ∈  (  0  ;  1  )Our initial limit      lim          N      →      ∞            (    2          N      +      1            −              ∑              n        =        1            N              1              n              )    =      lim          N      →      ∞            (    2          N            −              ∑              n        =        1            N              1              n              )    =  −  ζ            (            1    2              )        =  1.46035...

I ran into this limit while evaluating an integral. <munder> <mo movablelimits="true" form="

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New Questions in Calculus 1