If S = 1^i+2^i+3^i+…+n^i where 'i' is imaginary unit and n -> inf?

Picking up from where deighton left off,

n^i = exp(i*ln(n)) = cos(ln n) + i sin(ln n)

S = ∑ cos(ln n) + i sin(ln n)

for n = 1 to ∞

= Re + i* Im

Both the real and imaginary sums can be viewed as Reimann sums for the respective integrals:

Re = ∫{x=1..n} cos(ln(x)) dx = x/2 * [cos(lnx) + sin(lnx)]

and

Im = ∫{x=1..n} sin(ln(x)) dx = x/2 * [sin(lnx) – cos(lnx)]

Applying the limits

Re = n/2 * [cos(ln(n)) + sin(ln(n))] – 1/2

Im = n/2 * [sin(ln(n)) – cos(ln(n))] + 1/2

|S|^2 = n^2 /2 – n*cos(ln(n)) + 1/2

after simplification

As n –> ∞, the last two terms become negligible compared to the first on the RHS

ie |S|^2 –> n^2 /2

|S| –> n/√2

|S| – n/√2 –> 0

82

here’s what I got

x is an integer

x^i = exp(log(x^i)) = exp(i log(x)) = cos(log(x)) + i sin(log(x))

I can’t see that it can have a limit, basically the summation keeps adding a complex number vector always of length 1 who’s value will change over time, but for periods of time it will be doubling back on itself.

What exactly is |S| defined as? I presumed sqrt(a^2 + b^2) where a+bi is the complex number resulting from the summatiion

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issue with further answer.

I’m really interested in this, so I’m not trying to pick fault, is a Rieman sum not an approximation? The equating of the summation to an integral is surely an approximation, or are you perhaps suggesting that the approximation tends to equality in the limit? But I’m having a hard time seeing that.

This expression will never actually converge to zero. It will keep oscillating between ~ 0.78 and -0.78. This is my observation from computer generated data.

The expression Dr D derived does not converge to 0 either. It oscillates between 1/sqrt(2) and -1/sqrt(2)

(But I didn’t give anyone a thumbsdown! Just wanted to say this since it may look like it now that I am disagreeing)

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Also, the length of the cycle of both oscillations grow exponentially.

I see that Dr D has beat me to this. His answer is correct. The limit (for positive integer n, that is) is 0.