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8 votes
Accepted

$\frac{1}{x} + \frac{1}{x-1} + \ldots + \frac{1}{x-n} = 0$ has only one root $(0;1)$

If it has two roots then its derivative will have to vanish at some point (by Rolle's Theorem). But the derivative is striclty negative at every point of $(0,1)$.
geetha290krm's user avatar
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4 votes

$\frac{1}{x} + \frac{1}{x-1} + \ldots + \frac{1}{x-n} = 0$ has only one root $(0;1)$

Using calculus, $\frac{d}{dx}\frac{1}{x-\alpha}=-\frac{1}{(x-\alpha)^2}$, so the derivative of the original function is always negative.Therefore, it can have only one real zero in the region $(0,1)$ ...
Zima's user avatar
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3 votes

$\frac{1}{x} + \frac{1}{x-1} + \ldots + \frac{1}{x-n} = 0$ has only one root $(0;1)$

As Zima wrote in comment, you proved there is at least one root. To prove there is at most one root, note that every term monotonically decreases for $x \in (0, 1)$ - so the sum also monotonically ...
mihaild's user avatar
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2 votes

How to find principal value of the cubic root?

According to a comment on the similar question, the first non-real root as you traverse counter clockwise from the positive real axis is chosen by Mathematica.
Zack Fisher's user avatar
1 vote

$\frac{1}{x} + \frac{1}{x-1} + \ldots + \frac{1}{x-n} = 0$ has only one root $(0;1)$

As said by others, the function is strictly decreasing. But it has a vertical asymptote with a change of sign at every non-negative integer. Hence it has a single root in every interval $(m,m+1)$ from ...
Yves Daoust's user avatar

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