Prove that for every $n\in \mathbb{N}^{+}$, there exist a unique $x_{n}\in[\frac{2}{3},1]$ such that $f_{n}(x_{n})=0$

Question

Let $f_{n}(x)=-1+x+\dfrac{x^2}{2^2}+\dfrac{x^3}{3^2}+\cdots+\dfrac{x^n}{n^2}$,

(1) Prove that for every $n\in \mathbb{N}^{+}$, then there exist unique $x_{n}\in[\frac{2}{3},1]$ such that $f_{n}(x_{n})=0$

(2) Show that the sequence $(x_{n})$ of (1) is such that $$0<x_{n}-x_{n+p}<\dfrac{1}{n}$$ for all $p\in \mathbb{N}$.

(3):$x_{n}=A+\dfrac{B}{n}+\dfrac{C}{n^2}+O(\dfrac{1}{n^2})$, find $A,B,C$?

My attempts :

For $(1)$, I have prove it : $$f_{n}(1)=-1+1+\dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2}>0,$$ \begin{align} f_{n}(\frac{2}{3})&=-1+\dfrac{2}{3}+\cdots+\dfrac{(\dfrac{2}{3})^n}{n^2}\\ &\le-\dfrac{1}{3}+\dfrac{1}{4}\sum_{k=2}^{n}\left(\dfrac{2}{3}\right)^k\\ &=-\dfrac{1}{3}\cdot\left(\dfrac{2}{3}\right)^{n-1}<0 \end{align}

and $$f'_{n}(x)=1+\dfrac{x}{2}+\cdots+\dfrac{x^{n-1}}{n}>0$$

But for $(2)$, I have methods:

since $$x>0, f_{n+1}(x)=f_{n}(x)+\dfrac{x^{n+1}}{(n+1)^2}>f_{n}(x)$$ so $$f_{n+1}(x_{n})>f_{n}(x_{n})=f_{n+1}(x_{n+1})=0$$ since $f_{n+1}(x)$ is increasing, so $x_{n+1}<x_{n}$

since $$f_{n}(x_{n})=-1+x_{n}+\dfrac{x^2_{n}}{2^2}+\cdots+\dfrac{x^n_{n}}{n^2}=0$$

$$f_{n+p}(x_{n})=-1+x_{n+p}+\dfrac{x^2_{n+p}}{2^2}+\cdots+\dfrac{x^{n+p}_{n+p}}{(n+p)^2}=0$$ and use $0<x_{n+p}<x_{n}\le 1$ we have

\begin{align} x_{n}-x_{n+p}&=\sum_{k=2}^{n}\dfrac{x^k_{n+p}-x^k_{n}}{k^2}+\sum_{k=n+1}^{n+p}\dfrac{x^k_{n+p}}{k^2}\le\sum_{k=n+1}^{n+p}\dfrac{x^k_{n+p}}{k^2}\\ &\le\sum_{k=n+1}^{n+p}\dfrac{1}{k^2}<\sum_{k=n+1}^{n+p}\dfrac{1}{k(k-1)}=\dfrac{1}{n}-\dfrac{1}{n+p}<\dfrac{1}{n} \end{align}

But for part $(3)$ I can't prove it,Thank you everyone can help,and for part $(2)$ have other nice methods?

Answer 1

$x_{n}-x_{n+p}>0$ trivial, since $f_{n+p}(x_{n})>f_{n}(x_{n})=0$ and these $f$ always increase.

For the other inequality, I couldn't find a neat way to do it, so pardon some of the messy calculation. The idea is to use mean value theorem:

If $n\leq 3$ then trivial. So assume $n\geq 4$:

Now a bit of number crunching go in. You could use a Ferrari's quartic formula, or you could show that $f_{4}(\frac{4}{5})>0$. Either way you get that $x_{4}<\frac{4}{5}$ and thus $x_{n}<\frac{4}{5}$ for all $n\geq 4$.

Now we have $n(\frac{(x_{n})^{n+1}}{(n+1)^{2}}+\frac{(x_{n})^{n+2}}{(n+2)^{2}}+...+\frac{(x_{n})^{n+p}}{(n+p)^{2}})<\frac{(x_{n})^{n+1}}{n+1}+\frac{(x_{n})^{n+2}}{n+2}+...+\frac{(x_{n})^{n+p}}{n+p}<\frac{(x_{n})^{n+1}}{n+1}+\frac{(x_{n})^{n+2}}{n+1}+...+\frac{(x_{n})^{n+p}}{n+1}<\frac{(x_{n})^{n+1}}{n+1}(1+x_{n}+(x_{n})^{2}+...)=\frac{(x_{n})^{n+1}}{n+1}\frac{1}{1-x_{n}}$.

Use the earlier bound $n\geq 4,x_{n}<\frac{4}{5}$ give $\frac{(x_{n})^{n+1}}{n+1}\frac{1}{1-x_{n}}<\frac{(\frac{4}{5})^{5}}{5}\frac{1}{1-\frac{4}{5}}<1$

Notice that $f'_{n+p}(x)=1+\frac{x}{2}+...+\frac{x^{n+p-1}}{n+p}>1$ so $f'_{n+p}(x)>n(\frac{(x_{n})^{n+1}}{(n+1)^{2}}+\frac{(x_{n})^{n+2}}{(n+2)^{2}}+...+\frac{(x_{n})^{n+p}}{(n+p)^{2}})=nf_{n}(x_{n})+n(\frac{(x_{n})^{n+1}}{(n+1)^{2}}+\frac{(x_{n})^{n+2}}{(n+2)^{2}}+...+\frac{(x_{n})^{n+p}}{(n+p)^{2}})=n(-1+x_{n}+\frac{(x_{n})^{2}}{2^{2}}+...+\frac{(x_{n})^{n+p}}{(n+p)^{2}})=nf_{n+p}(x_{n})$

Now if $x_{n}-x_{n+p}\geq\frac{1}{n}$ then by applying mean value theorem there exist some $x\in[x_{n+p},x_{n}]$ such that $f'_{n+p}(x)=\frac{f_{n+p}(x_{n})-f_{n+p}(x_{n+p})}{x_{n}-x_{n+p}}=\frac{f_{n+p}(x_{n})}{x_{n}-x_{n+p}}\leq\frac{f_{n+p}(x_{n})}{(\frac{1}{n})}=nf_{n+p}(x_{n})$ which contradict the inequality just derived above.

Answer 2

This is not really a solution for $(3)$, but according to numerical evidence, $$\tag1A=\lim_{n\to\infty}x_n\approx 0.76154294453204558806805187241022287341383\ldots$$ as estimated from $x_{100,000}$. By $(2)$, only the first five digits need to be correct. However, computing with high precision, I find $|x_{100,000}-x_{1,000}|<10^{-120}$, so this suggests that the convergence is quite rapid, i.e. $B=C=0$. Rather I would guess that $|x_n-A|\sim \frac{A^n}{n^2}$.

Answer 3

The $f_n$ converge to $f(x) = f_\infty(x) = -1 + \displaystyle \sum_{n=1}^{\infty} \frac{x^k}{k^2} $ = Li_2$(x)-1$ (the dilogarithm, minus $1$) with remainder terms $R_n(x) = \displaystyle \sum_{k=n}^{\infty} \frac{x^k}{k^2}$. Then:

1. $x_n$ is a decreasing sequence that converges to $A \in (0,1)$ with $f(A)=0$, the inverse dilogarithm of $1$.

2. The inverse of $f$ has a power series expansion near $0$ with real coefficients and a positive radius of convergence (due to analyticity of $f$ for $|z|<1$ and nonzero derivative of $f$ on $(0,1)$), $$ f^{-1}(z)= A + \beta z + \gamma z^2 + \dots$$ where $\beta$ and $\gamma$ can be calculated from $A$, $f'(A) = -\log(1-A)$, and $f''(A)=\frac{1}{1-A}$ by formal inversion of series.

3. $f(x_{n}) = R_{n+1}(x)$ so that $x_n = f^{-1}(R_{n+1}(x_n))$ once $n$ is large enough that $R_{n+1}$ is in the circle of convergence.

4. $R_p(x)$ is between $\frac{x^p}{p^2}$ and $\frac{x^p}{p^2} (1-x)^{-1}$ by comparing terms and is within a factor of $1 + O(\frac{1}{p})$ of the upper bound for constant positive $x < 1$

5. From the formula in (3), $\quad x_{n-1} $ differs from $A$ by $O(x_n^n / {n^2}$) which is much smaller than a polynomial in $\frac{1}{n}$. This difference is dominated, for large $n$ and any positive $u$, by $(A+u)^n / {n^2}$ which implies that the asymptotic expansion of $x_n$ in powers of $\frac{1}{n}$ with constant coefficients is $A + \frac{0}{n} + \frac{0}{n^2} + \frac{0}{n^3} + \cdots$. Better than that, to any finite number of terms in the power series of $f^{-1}(z)$ there is no asymptotic difference between use of $x_n$ and $A$ in the formulas, and the first two terms of the asymptotics are $$x_{n} = f^{-1}(R_{n+1}(A)) = A + \beta \frac{A^{n+1}}{(1-A)(n+1)^2} + O(R^2) = A + (\frac{\beta A}{1-A}) \frac{A^n}{n^2}$$ plus terms of higher order in $A^n$ or $(1/n))$.

Conclusions:

• The answer to the question as written is $B=C=0$ and $A = ($ inverse_ dilogarithm of $1$) $= 0.761542944532045588068051872410222873413830030301381136443779....$, but the true magnitude of $(x_n - A)$ is about $\frac{A^n}{n^2}$. Exactly as foreseen in the earlier answer from Hagen von Eitzen.

• $\beta = \frac{1}{\log(\frac{1}{1-A})}$

• going to order $\gamma z^2$ looks challenging.

Answer 4

Clearly $ (x_n) $ is monotone decreasing and bounded below by 2/3. Therefore $ (x_n) $ converges. Since we are in $ \mathbb{R} $, this means that $ (x_n) $ is Cauchy. I'll let you look up the definition of a Cauchy sequence, but at this point, what you need to do is fix an $ N \in \mathbb{N} $, and let $ \epsilon = 1/N $. You know that there exists an $ M \in \mathbb{N} $ such that when $ n \geq M $, $ |x_n - x_{n+p}| = x_n-x_{n+p} < 1/N $, so it remains to prove that the $ M $ that works is equal to the $ N $ you fixed.