What do temporal and spatial summations have in common?

Exponetial summations?

• an = e^ (n/(n+1)) What would the sum to N, SN, for this expression from n=1 to n=N. (I.e. for ∑n between n=1 and n=N SN = N(N+1)/2 What about ∑e^ (n/(n+1)) between n=1 and N ? Thanks. (Ps - does this follow geometric version of sn in anyway? = (1-q^n)/(1-q))

• Answer:

  It's a very interesting question. this is what I came up with: e^[n/(n+1)] = e^[(n+1-1)/(n+1)] = e^[{(n+1)/(n+1)}-{1/(n+1)}] = e^[1 - {1/(n+1)}] = {e^1} / {e^[1/(n+1)]} = e / {e^[1/(n+1)]} a_1 = e / {e^[1/2]} a_2 = e / {e^[1/3]} a_3 = e / {e^[1/4]} a_2 / a_1 = a_3 / a_2 gives you that 1/6 = 1/12 ==> a_n IS NOT A GEOMETRIC SERIES Let's analyze e^[1/(n+1)] = radical_of_order_(n+1)_of_(e). As n-->infinity the radical_of_order_n-->1 ==> radical_of_order_(n+1)-->1 1 < e^[1/(n+1)] <= e^(1/2) for any "n >= 1" ==> 1/1 > 1 / {e^[1/(n+1)]} >= 1 / {e^(1/2)} 1/{e^(1/2)} <= 1 / {e^[1/(n+1)]} < 1. . . . . . . .|*e e/{e^(1/2)} <= e / {e^[1/(n+1)]} < e e^(1/2) <= a_n < e sqrt(e) <= a_n < e. sqrt(e) <= a_1 < e sqrt(e) <= a_2 < e sqrt(e) <= a_3 < e .......................... sqrt(e) <= a_(N-1) < e sqrt(e) <= a_N < e ------------------------------------ + N*sqrt(e) <= ∑{a_n} < N*e It's the best I could do for now. I also tried to write every e^[1/(n+1)] as a Taylor series but it got really messy so I aborted. Best regards! And let's hope someone will find something better than me.