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Let $x_n \in (0;1)$ be a positive real root of this function: $$f_n(x) = \frac{1}{x} + \frac{1}{x-1}+\ldots+\frac{1}{x-n}$$ with positive integer $n \geq 2$

Find $\lim x_n$

I claimed that $f_n(x) = 0$ has only one positive real root $x_n\in(0;1)$ for each positive integer $n$

From the equation: $$\frac{1}{x_n} = \frac{1}{1-x_n} + \frac{1}{2-x_n} + \ldots + \frac{1}{n-x_n} > \frac{1}{1} + \frac{1}{2} + \ldots + \frac{1}{n} > \frac{1}{1.2} + \frac{1}{2.3} + \ldots + \frac{1}{n(n+1)} = \frac{n+1}{n}$$

Thus:

$$0 < x_n < \frac{n}{n+1}$$

But the RHS can't apply Squeeze Theorem, anyone help me please!

Thomas Andrews
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Lục Trường Phát
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2 Answers2

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As you proved already $$\frac{1}{x_n} > \sum_{i=1}^n \frac{1}{i}$$ By knowing that $\sum_{i=1}^n \frac{1}{i} \xrightarrow{n\to +\infty} +\infty$: $$x_n < \frac{1}{\sum_{i=1}^n \frac{1}{i}}\xrightarrow{n\to +\infty} 0$$ We conclude that $\lim x_n = 0$.

NN2
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One can obtain concrete upper and lower bounds for the $x_n$ and so determine their asymptotics.

For the upper bound we proceed as in NN2's answer: $$ \frac{1}{x_n} > \sum_{k=1}^n \frac 1k > \int_1^{n+1} \frac{dt}{t} = \ln(n+1) > \ln(n) $$ so that $ x_n < \frac{1}{\ln(n)}$. In particular is $0 < x_n < 1/2$ for all $n \ge 8$, and that can be used to compute a lower bound in a similar way: $$ \frac{1}{x_n} < 2 + \sum_{k=2}^n \frac 1{k-1/2} < 2 + \int_{1/2}^{n-1/2} \frac{dt}{t} = 2 + \ln(n-1/2)+ \ln(2) < 3 + \ln(n) $$ so that $ x_n > \frac{1}{3 + \ln(n)}$. Together we have $$ \frac{1}{3 + \ln(n)} < x_n < \frac{1}{\ln(n)} $$ for all $n \ge 8$, and in particular $x_n \cdot \ln(n) \to 1$.

Martin R
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