What's the formula to solve summation of logarithms?I'm studying summation. Everything I know so far is that: ∑ i = 1 n k = n ( n + 1 ) 2 ∑ i = 1 n k 2 = n ( n + 1 ) ( 2 n + 1 ) 6 ∑ i = 1 n k 3 = n 2 ( n + 1 ) 2 4 Unfortunately, I can't find neither on my book nor on the internet what the result of: ∑ i = 1 n log ⁡ i ∑ i = 1 n ln ⁡ iis.Can you help me out?

Question

What&#039;s the formula to solve summation of logarithms?I&#039;m studying summation. Everything I know so far is that:      ∑          i      =      1        n       k  =            n      (      n      +      1      )        2           ∑          i      =      1              n               k    2    =            n      (      n      +      1      )      (      2      n      +      1      )        6           ∑          i      =      1              n               k    3    =                    n        2            (      n      +      1              )        2              4     Unfortunately, I can&#039;t find neither on my book nor on the internet what the result of:      ∑          i      =      1        n    log  ⁡  i      ∑          i      =      1        n    ln  ⁡  iis.Can you help me out?

Alexis Fields · Accepted Answer

By using the fact that  log  ⁡  a  +  log  ⁡  b  =  log  ⁡  a  bthen      ∑    n    log  ⁡  i  =  log  ⁡  (  n  !  )      ∑    n    ln  ⁡  i  =  ln  ⁡  (  n  !  )

Ciara Mcdaniel · Accepted Answer

Since   log  ⁡  (  A  )  +  log  ⁡  (  B  )  =  log  ⁡  (  A  B  ), then       ∑          i      =      1        n    log  ⁡  (  i  )  =  log  ⁡  (  n  !  ). I&#039;m not sure if this helps a lot since you have changed a summation of   n terms into a product of   n factors, but it&#039;s something.