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Let $n$ be a positive integer. Consider the function

$$ f_n(x)=\frac{1}{n}\bigg(\sum_{k=0}^{n-1}\frac{1}{x+\frac{k}{n}}\bigg) $$

Then $f_n$ is a decreasing bijection $]0,+\infty[ \to ]0,+\infty[$. So the equation $f_n(x)=1$ has a unique positive solution ; let us call it $x_n$. Also put $y_n=x_n-\frac{1}{2n}$.

Is it true that $(x_n)$ and $(y_n)$ are adjacent sequences, with $(x_n)$ decreasing and $(y_n)$ increasing ? What is the limit of $(x_n)$ and $(y_n)$ ?

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Limit $1/(\mathrm e-1)$, presumably. –  Did May 2 '12 at 6:38
    
@Blender: $]a,b[$ is another notation for the open interval $(a,b)$. –  Brian M. Scott May 2 '12 at 6:59
    
Didier : The integral-sum comparison yields $1-\frac{1}{nx_n} \leq \ln\big(1+\frac{1}{x_n}\big) \leq 1$. So, if $nx_n \to +\infty$, then $(x_n) \to \frac{1}{e-1}$ indeed. But how would we know that $nx_n \to +\infty$ ? –  Ewan Delanoy May 2 '12 at 7:02

2 Answers 2

up vote 1 down vote accepted

Since $\frac1z=\int\limits_0^1s^{z-1}\mathrm ds$ for every $z\gt0$, $$ f_n(x)=\frac1n\sum_{k=0}^{n-1}\int_0^1s^{k/n}\cdot s^{x-1}\mathrm ds=\int_0^1\frac{1-s}{h_n(s)}\cdot s^{x-1}\mathrm ds, $$ with $$ h_n(s)=n(1-s^{1/n}). $$ For every $c$, the derivative of the function $u\mapsto(1-\mathrm e^{-cu})/u$ is proportional to $1+cu-\mathrm e^{cu}\leqslant0$. Applying this to $u=1/n$ and $c=-\log(s)\geqslant0$ for every $s$ in $(0,1)$, one sees that the sequence $(h_n)_n$ is nondecreasing, hence the sequence $(f_n)_n$ is nonincreasing.

Now, for every $n$, $f_{n+1}(x_n)\leqslant f_n(x_n)=1=f_{n+1}(x_{n+1})$ and the function $f_{n+1}$ is nonincreasing hence $x_{n+1}\leqslant x_n$.

Likewise, consider the functions $$ g_n(y)=\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{y+\frac1{2n}+\frac{k}{n}}. $$ The integral representation used above nowyields $$ g_n(y)=\int_0^1\frac{1-s}{k_n(s)}\cdot s^{y-1}\mathrm ds, $$ with $$ k_n(s)=h_n(s)/s^{1/2n}=n(s^{-1/2n}-s^{1/2n}). $$ The derivative of the function $u\mapsto(\mathrm e^{cu/2}-\mathrm e^{-cu/2})/u$ is proportional to $\ell(u)$ with $$ \ell(u)=cu(\mathrm e^{cu}+1)-\mathrm e^{cu}+1. $$ Now, $\ell(0)=0$ and $\ell'(u)=c(1+cu\mathrm e^{cu})$, thus, for every $c\geqslant0$, $\ell'(u)\geqslant0$ for every $u\geqslant0$, and in particular $\ell(u)\geqslant0$, hence the function $u\mapsto(\mathrm e^{cu/2}-\mathrm e^{-cu/2})/u$ is nondecreasing on $u\geqslant0$.

Applying this to $u=1/n$ and $c=-\log(s)\geqslant0$ for every $s$ in $(0,1)$, one sees that the sequence $(k_n)_n$ is nonincreasing. Thus, the sequence $(g_n)_n$ is nondecreasing. The argument used for $(x_n)$ then shows that $(y_n)$ is nondecreasing.

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Define $g_n(t)=\frac{1}{x_n+t}$ for $t\in [0,1]$. Then $g_n$ is decreasing, so that for $t\in [\frac{k}{n},\frac{k+1}{n}]$ we have $g_n(\frac{k+1}{n}) \leq g_n(t) \leq g_n(\frac{k}{n})$. Integrating, we obtain $\frac{1}{n} \frac{1}{x_n+\frac{k+1}{n}} \leq \int_{\frac{k}{n}}^{\frac{k+1}{n}} g_n(t) \leq \frac{1}{n} \frac{1}{x_n+\frac{k}{n}}$ . Summing from $k=0$ to $k=n-1$, we deduce $$ f_n(x_n)+\frac{1}{n} \big(\frac{1}{x_n+1}-\frac{1}{x_n}\big) \leq \int_{0}^{1} g_n(t) \leq f_n(x_n) $$ so that $$ 1-\frac{1}{nx_n(x_n+1)} \leq {\sf ln} \big(1+\frac{1}{x_n}\big) \leq 1 $$ From the rightmost inequality we deduce that $x_n \geq \frac{1}{e-1}$. It follows that the left-hand side $1-\frac{1}{nx_n(x_n+1)} $ converges to $1$, so that ${\sf ln} \big(1+\frac{1}{x_n}\big)$ converges to $1$, and hence $(x_n)$ converges to $\frac{1}{1-e}$.

I don't know how to show the monotonicity properties of $(x_n)$ and $(y_n)$ though.

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