Verified answer

No, it does not necessarily involve Fourier series. Just rewrite the sine as the imaginary part of exp(in):

Im Σ [n=1 to +∞] exp(in)/n.

Note that the exp part can be written (e^i)^n. Thus our series is "almost" a geometrical one. Actually, it is just a tiny bit harder itself, but the main problem comes with the complex numbers.

To compute the generalized form of the same problem,

f(x) = Σ [n=1 to +∞] x^n/n,

note that the first derivative, when computed term-wise, is

f'(x) = Σ [n=1 to +∞] x^(n-1) = Σ [m=0 to +∞] x^m = 1/(1-x).

You surely know that this result is valid for real x such that |x|<1, that is, -1 < x < 1. From the theory of power series, the validity can be extended to all complex x's such that |x|<1, which is the unit disk.

We can integrate f'(x) back to f(x) to get a formula

f(x) = ∫ 1/(1-x) dx = -ln (1-x) + C.

The integration constant C can be determined from the condition

f(0) = Σ 0^n/n = 0

f(0) = -ln(1-0) + C = C

=> C = 0.

There is one more glitch in the above formula. As opposed to real numbers, the "ln" function is multi-valued on C. However, we chose the "principal branch" by assuming that ln 1 = 0.

From the theory again, it follows that the integrated series f(x) can converge on a slightly larger domain than f'(x): it can converge in some additional points of its boundary, where the derivative is not defined. This is very good for us as in our case, x equals e^i for which |x| = 1 (thus x lies on the boundary of the unit disk—the unit circle).

The actual domain of convergence of the series defining f(x) can be found to be {x in C | |x| <= 1 and x ≠ 1}. The analytical expression given above is valid on this whole set without any further corrections. Hence, we can write

f(e^i) = Σ [n=1 to +∞] exp(in)/n = -ln(1-e^i).

Now we have completed the summation and the task is just to find the imaginary part of this number. Recall that the imaginary part of a complex logarithm is the argument (phase) of the given complex number. To find it, we use the following trick:

1 - e^i = e^(i/2) * (e^(-i/2) - e^(i/2))

The second term is -2i times the sine of 1/2, which in turn is just some positive real number:

1 - e^i = e^(i/2) * (-i) * (2 sin 1/2) = (2 sin 1/2) * e^(i/2 - πi/2) = (2 sin 1/2) * e^((1-π)/2*i).

The argument of 1-e^i can be then simply identified as (1-π)/2.

Finally, together with the "-" before the logarithm, the imaginary part is

Σ [n=1 to +∞] sin(n)/n = Im (-ln(1-e^i)) = -(1-π)/2 = (π-1)/2.

I hope this helps you! In case of any questions please state them in the Additional details.

I found this in a discussion elsewhere. The value of the sum converges to (π–1)/2. The analysis is pretty advanced. Here is a quote from the discussion. You need considerable advanced analysis to do this:

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The value for the sum follows from

1. Fourier analysis: take x on (-pi,pi] and compute the Fourier coefficients

and use that the Fourier series converges for x = pi/2.

2. Complex numbers: -log(1-x) = Sum_{n>=1} [x^n / n], so

we need the imaginary part of -log(1-e^i).....

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Apparently it is an consequence of a Fourier Series expansion. Not for novices.