How can I prove that the following function is increasing in $x$: $$\sum_{i=1}^{\infty} x (1-x) ^ {i-1} \log \left (1+ \mu (1-x)^{i-1} \right)$$
where $\mu$ is any non-negative number and $x$ is in $[0,1]$?
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$\begingroup$ You can differentiate with respect to $x$ through the sum. $\endgroup$– Simply Beautiful ArtCommented Jul 26, 2016 at 22:27
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$\begingroup$ Is false. If x=0 or x=1, the sum is 0. If is increasing, must be 0 for all x, abs $\endgroup$– Martín Vacas VignoloCommented Jul 27, 2016 at 1:15
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$\begingroup$ No, the sum at x=1 is the maximum. Only the first term of the summation will be non-zero. So for x=0 it is zero, but for x=1 is equal to $\log(1+\mu)$ (which is the maximum value of this summation) $\endgroup$– PatrikCommented Jul 27, 2016 at 13:30
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$\begingroup$ Differentiating will not take us anywhere, since it will be an infinite summation with mixed terms (with logarithm and fractions), $\endgroup$– PatrikCommented Jul 27, 2016 at 13:31
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$\begingroup$ @Patrik if x=1 the term (1-x) is 0, then the sum is 0 $\endgroup$– Martín Vacas VignoloCommented Jul 28, 2016 at 4:18
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