Rate of convergence of the given sequence

Question

Consider the sequence: $x_n=\frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{n+n}$. Find the rate of convergence of the sequence $\{x_n\}$.

For this, first we find the limit of the sequence $\{x_n\}$.

We have,

\begin{align*} \lim_n x_n&=\lim_n\left[\frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{n+n}\right]\\ &=\lim_n\frac 1n \sum_{r=1}^n \frac{1}{1+\frac rn}\\ &=\int_0^1 \frac{1}{1+x}\,dx\\ &=\log(2) \end{align*}

To find the rate of convergence we have to find out the value $p$ so that $\displaystyle \lim_{n\to \infty} \frac{|x_{n+1}-l|}{|x_n-l|^p}$ exists non-zero finitely.

I am unable to find out this limit. Any help please ?

Answer 1

All of what you want lies in the rectangle method, let us calculate

$$ \ln(2) - x= \sum_{k=1}^n \int_{\frac{k-1}{n}}^\frac{k}{n} \frac{1}{1 +x} - \frac{1}{1 +\frac{k}{n}} dx \geq 0$$ A quick study of the derivative would give us quickly that $|\ln(2) - x_n| \leqslant M/n$ for some $M$, but we need to show that we can't do better...

$$ \int_{\frac{k-1}{n}}^\frac{k}{n} \frac{1}{1 +x} - \frac{1}{1 +\frac{k}{n}} dx \geq \int_{\frac{k-1}{n}}^\frac{k- 1/2}{n} \frac{1}{1 +x} - \frac{1}{1 +\frac{k}{n}} \geq \frac{1}{2n} \left(\frac{1}{1 + \frac{2k-1}{2n}} - \frac{1}{1 + \frac{k}{n}}\right) \\ = \frac{1}{4n^2}\cdot\frac{1}{(1 + \frac{k}{n})(1 + \frac{2k-2}{2n})}\geq \frac{1}{16n^2}$$ So we get $$\ln(2) - x \geq n \cdot \frac{1}{16n^2} = \frac{1}{16n}$$ This gives you directly that $p = 1$.

Answer 2

For $f\in C^1[1,2]$ there holds $$\left |{1\over n}\sum_{k=n+1}^{2n}f\left ({k\over n}\right)-\int\limits_1^2f(x)\,dx\right | \le {M\over n}\quad (*)$$ where $M=\displaystyle \max_{1\le x\le 2}|f'(x)|.$ We have $$\log (1+x)=x-{x^2\over 2}+O(x^3),\quad 0\le x<1$$ Hence $$\sum_{k=n+1}^{2n}{1\over k} =\sum_{k=n+1}^{2n}\log\left (1+{1\over k}\right )+{1\over 2}\sum_{k=n+1}^{2n}{1\over k^2}+O(n^{-2}) $$ Next $$\sum_{k=n+1}^{2n}\log\left (1+{1\over k}\right )=\log{2n+1\over n+1}\\ =\log 2+\log \left (1-{1\over 2n+2}\right )\\ = \log 2-{1\over 2n+2}+O(n^{-2})=\log 2-{1\over 2n}+O(n^{-2})$$ Applying $(*)$ we get $${1\over 2}\sum_{k=n+1}^{2n}{1\over k^2}={1\over 2n^2}\sum_{k=n+1}^{2n}{1\over ({k\over n})^2} ={1\over 2n} \int\limits_1^2{1\over x^2}\,dx+O(n^{-2})\\ ={1\over 4n}+O(n^{-2})$$ Summarizing $$x_n-\log 2= \sum_{k=n+1}^{2n}\log\left (1+{1\over k}\right )-\log 2=-{1\over 4n}+O(n^{-2})$$ Thus $p=1.$

Answer 3

If you use harmonic numbers $$x_n=H_{2 n}-H_n$$ Now, when $p$ is large $$H_p=\log (p)+\gamma +\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^4}\right)$$ whic makes $$x_n=\log (2)-\frac{1}{4 n}+\frac{1}{16 n^2}+O\left(\frac{1}{n^4}\right)=\log (2)-\frac{1}{4 n}+O\left(\frac{1}{n^2}\right)$$