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TopNewestSo, to solve this problem, a substitution must be made. Let \[y_i = \frac{1}{2} - x_i\]. Therefore \[0 < y_i < \frac{1}{2}\] for all \[i\]. Also of note, \[\frac{1}{2} + x_i = 1 - y_i\] is easily verified. By this definition, \[a_1 = 1 - y_1, a_2 = (1 - y_2) y_1, a_3=(1 - y_3) y_2 y_1,...\] or in general, \[a_i = (1 - y_i) \displaystyle \prod_{i=1}^{i - 1} y_i = \displaystyle \prod_{i=1}^{i - 1} y_i - \displaystyle \prod_{i=1}^i y_i\]. I'll let the reader explore why the forth statement given isn't necessary. So from this, \[\displaystyle \sum_{k=1}^\infty ka_k = 1 - y_1 + 2y_1 - 2y_1y_2 +3y_1y_2 - 3y_1y_2y_3 + 4y_1y_2y_3y_4 \ldots = 1 + y_1 + y_1y_2 + y_1y_2y_3 + \ldots\]. The only way to maximize the sum is to have every \[y_i = \frac{1}{2}\]. Doing so produces 2, and thus equality with the final statement. All other \[y_i\] are less than that. For more rigor, one can show that the previous sum is \[1 + y_1(1 + y_2(1 + y_3(1 + \ldots\], each parentheses of which must be less than 2. The answer becomes much more apparent when you make the intial substitution. – Bob Krueger · 4 years, 3 months ago

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