I would like to compute: ∑ i = 1 ∞ ∑ j = 1 ∞ ( − 1 ) i + j i + j

Question

I would like to compute:      ∑          i      =      1              ∞            ∑          j      =      1              ∞                  (      −      1              )                  i          +          j                            i      +      j

Cahokiavv · Accepted Answer

How about:  −      ∫    0    1    (  −  x      )          i      +      j      −      1          d  x  =            (      −      1              )                  i          +          j                            i      +      j      then for each   x  ∈  (  0  ,  1  ) we have      ∑          i      =      1        ∞        ∑          j      =      1        ∞    −  (  −  x      )          i      +      j      −      1        =      x          (      x      +      1              )        2            and integrate      ∫    0    1        x          (      x      +      1              )        2                d  x  =  log  ⁡  2  −      1    2    ≈  0.193147

sedeln5w · Accepted Answer

Let       S    n    =      ∑          i      ⩾      1      ,      j      ⩾      1            (                  (        −        1                  )                      i            +            j                                      i        +        j                            1                    i        +        j        ⩽        n        +        1              )  . Then      S    n    =      ∑          i      =      1        n        ∑          j      =      1              n      +      1      −      i                  (      −      1              )                  i          +          j                            i      +      j        =      ∑          i      =      1        n        ∑          j      =      1              n      +      1      −      i            ∑          m      =      2              n      +      1            δ          i      +      j      ,      m                  (      −      1              )                  i          +          j                            i      +      j        =      ∑          m      =      2              n      +      1            ∑          i      =      1        n        ∑          j      =      1              n      +      1      −      i            δ          i      +      j      ,      m                  (      −      1              )                  m                      m    =      ∑          m      =      2              n      +      1        (  −  1      )    m              m      −      1        m    =      ∑          m      =      2              n      +      1        (  −  1      )    m    −      ∑          m      =      2              n      +      1                  (      −      1              )        m              m    =      1    2        (    1    −    (    −    1          )      n        )    +      ∑          m      =      1        n              (      −      1              )        m                    m      +      1      Notice that      lim          n      →      ∞            S          2      n        =      lim          n      →      ∞            ∑          m      =      1              2      n                  (      −      1              )        m                    m      +      1        =      lim          n      →      ∞            (    ln    ⁡    (    2    )    −    1    +          H              n        +        1                  /                2              −          H      n        )    =  ln  ⁡  (  2  )  −  1and      lim          n      →      ∞            S          2      n      +      1        =      lim          n      →      ∞            (    1    +          ∑              m        =        1                    2        n        +        1                            (        −        1                  )          m                            m        +        1              )    =      lim          n      →      ∞            (    1    +    ln    ⁡    (    2    )    −    1    +          H              n        +        1                  /                2              −          H              n        +        1              )    =  ln  ⁡  (  2  )Thus the sequence       S    n   does not converges as n increases, meaning that the original sum is not defined. The sequence       S    n   has Cesaro mean, though:      lim          n      →      ∞            1    n        ∑          m      =      1        n        S    m    =      1    2        (    ln    ⁡    (    2    )    +    (    ln    ⁡    (    2    )    −    1    )    )    =  ln  ⁡  (  2  )  −      1    2

I would like to compute: <munderover> ∑ <mrow class="MJX-TeXAtom-ORD">

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Answer & Explanation

New Questions in Calculus 2