Question

Question #be944

Answer 1

If we expand the summation as follows

`1/n sum_{k=0}^(n-1) e^{k/n} = 1/n { e^(0/n)+e^(1/n)+e^(2/n) + e^(3/n) + ... + e^((n-1)/n) }`
`" " = 1/n { underbrace(e^0+e^(1/n)+(e^(1/n))^2 + (e^(1/n))^3 + ... + (e^(1/n))^(n-1))_("n terms") }`

So you are correct, It is a GP with;

first term `a=e^0 (=1)` and
common ratio `r=e^(1/n)`

So Using the GP summation formula:

`S_n=a((1-r^n))/((1-r))`

to get:

`1/n sum_{k=0}^(n-1) e^{k/n} = 1/n1((1-(e^(1/n))^n))/((1-e^(1/n)))`
`" " = ((1-e^(n/n)))/(n(1-e^(1/n)))`
`" " = ((e^(n/n)-1))/(n(e^(1/n)-1))`
`" " = ((e^(n/n)-e^0))/(n(e^(1/n)-1))` (as `e^0=1`) QED