Let σ ( n ) = ∑ d ∣ n dthe sum of divisor function. I've deduced, but I don't know how prove directly σ ( n ) = ∑ m = 1 n ∑ k = 1 m ( − 1 ) n k cos ⁡ ( π n ( 2 − m ) k m ) .

Question

Let  σ  (  n  )  =      ∑          d      ∣      n        dthe sum of divisor function. I&#039;ve deduced, but I don&#039;t know how prove directly  σ  (  n  )  =      ∑          m      =      1              n            ∑          k      =      1              m        (  −  1      )          n      k        cos  ⁡      (                  π        n        (        2        −        m        )        k            m        )    .

delalbaef · Accepted Answer

Suppose we have the usual   σ  (  n  )  =      ∑          d              |            n        d and wish to show that  σ  (  n  )  =      ∑          m      =      1        n        ∑          k      =      1        m    (  −  1      )          n      k        cos  ⁡      (                  π        n        (        2        −        m        )        k            m        )    .This is      ∑          m      =      1        n        ∑          k      =      1        m    (  −  1      )          n      k        (  −  1      )          n      k        cos  ⁡      (                  2        π        n        k            m        )    =  ℜ      (          ∑              m        =        1            n              ∑              k        =        1            m              e              2        π        i        n        k                  /                m              )    .which is  ℜ      (          ∑              m        =        1            n              e              2        π        i        n                  /                m                    ∑              k        =        0                    m        −        1                    e              2        π        i        n        k                  /                m              )    =  ℜ      (          ∑              m        =        1            n              e              2        π        i        n                  /                m              ×    m    ×    [    [    m          |        n    ]    ]    )      =  ℜ      (          ∑              m                  |                n              m    )    =  σ  (  n  )as claimed.Here we have used the fact that      ∑          k      =      0              m      −      1            e          2      π      i      n      k              /            m        =                    e                  2          π          i          n                    −      1                      e                  2          π          i          n                      /                    m                    −      1        =  0when       e          2      π      i      n              /            m        ≠  1 and m otherwise which yields  m  ×  [  [  m      |    n  ]  ]  .

Let σ ( n ) = <munder> ∑ <mrow clas

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