# Tag Info

## Hot answers tagged roots

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 $(0,1)$.
• 39.2k

### $\frac{1}{x} + \frac{1}{x-1} + \ldots + \frac{1}{x-n} = 0$ 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 $(0,1)$ ...
• 3,392

### $\frac{1}{x} + \frac{1}{x-1} + \ldots + \frac{1}{x-n} = 0$ 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 ...
• 15.9k
### $\frac{1}{x} + \frac{1}{x-1} + \ldots + \frac{1}{x-n} = 0$ 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 ...