Joy of Mathematics

A limit

April 9, 2023April 13, 2023 Tolaso

Let $\theta \in (0, 1)$ . We define the sequence $\displaystyle{a_n = \left\{\begin{matrix} 0 & , & \left \lfloor n \theta \right \rfloor = \left \lfloor \left ( n-1 \right ) \theta \right \rfloor \\ 1 & , & \text{otherwise} \end{matrix}\right.}$ . Prove that

$\lim_{n \rightarrow +\infty} \frac{a_1 + a_2 + \cdots + a_n}{n} = \theta$

Sum of reciprocal sequence

March 10, 2020 Tolaso

Let $\{a_n\}_{n \in \mathbb{N}}$ be a sequence such that $a_1=3$ and

$a_{n+1} = a^2_{n}-2 \quad \text{forall} \;\; n \geq 2$

Evaluate the sum

$\mathcal{S} = \sum_{n=1}^{\infty} \prod_{k=1}^{n} \frac{1}{a_k}$

Solution

Let us consider the sequence

$b_n=\frac{a_n-\sqrt{a_n^2-4}}{2}$

and observe that

$\begin{align*} b_{n+1}&=\frac{a_{n+1}-\sqrt{a_{n+1}^2-4}}{2} \\ &=\frac{a_{n}^2-2-\sqrt{(a_{n}^2-2)^2-4}}{2} \\ &=\frac{a_{n} \left(a_n-\sqrt{a_n^2-4} \right)-2}{2} \\ &=a_nb_{n}-1 \end{align*}$

This in return means,

$b_n=\frac{1}{a_n}+\frac{b_{n+1}}{a_n}$

Thus,

$\begin{align*} b_1 &= \frac{1}{a_1} + \frac{b_2}{a_1} \\ & = \frac{1}{a_1} + \frac{1}{a_1} \left ( \frac{1}{a_2} + \frac{b_3}{a_2} \right ) \\ & = \frac{1}{a_1} + \frac{1}{a_1 a_2} \left ( \frac{1}{a_3} + \frac{b_4}{a_3} \right ) \\ &= \cdots \end{align*}$

which is the desired result. Therefore,

$\mathcal{S} = \sum_{n=1}^{\infty} \prod_{k=1}^{n} \frac{1}{y_k} = b_1 = \frac{3-\sqrt{5}}{2}$

Notes:

The recursive relation $a_{n+1} = a_n^2 + c$ has a closed form if-f $c=0$ or $c=-2$ . In our case it is:
$a_n = \alpha^{2^{n-1}}+\frac{1}{\alpha^{2^{n-1}}}$

where $\alpha = \frac{3 \pm \sqrt{5}}{2}$ .
The number $3-\sqrt{5}$ is known as the Grafting constant.
Under the same assumptions it holds that
$\lim_{n \rightarrow +\infty} \left( a_1 a_2 a_3 \dots a_n \right)^{\frac{1}{2^{n}}}=\frac{3+\sqrt{5}}{2}$

Limit of a sequence

April 6, 2019June 30, 2019 Tolaso

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a function such that $f(0)=0$ and $f$ is differentiable at $0$ . Let us set

$a_n =\sum_{k=1}^{n} f \left( \frac{k}{n^2} \right)$

Evaluate the limit $\lim \limits_{n \rightarrow +\infty} a_n$ .

Solution

Since $f$ is differentiable at $0$ , there is some $\varepsilon: x \mapsto \varepsilon(x)$ such that

$f(x) = f(0) + x f'(0) + x \varepsilon(x) \quad , \quad \varepsilon(0) =0$

and $\varepsilon$ is of course continuous.

Thus,

$\sum_{k=1}^{n} f\left ( \frac{k}{n^2} \right ) = \frac{f'(0)}{n} \sum_{k=1}^{n} \frac{k}{n} + \sum_{k=1}^{n} \frac{k}{n^2} \varepsilon \left ( \frac{k}{n^2} \right )$

Let $\epsilon>0$ . There exists $\delta>0$ such that $|x| \leq \delta$ which in return means that $|\varepsilon(x)| \leq \epsilon$ . Hence , for $n$ larger than $\frac{1}{\delta}+1$ it holds that

$\left| \sum_{k=1}^n \frac{k}{n^2}\varepsilon \left( \frac{k}{n^2} \right) \right|\leq \epsilon \frac {n}{n} = \epsilon$

On the other hand , the sum $\displaystyle \frac{1}{n} \sum_{k=1}^{n} \frac{k}{n}$ is a Riemann sum and converges to $\frac{1}{2}$ .

In conclusion,

$\lim_{n \rightarrow +\infty} a_n = \frac{f'(0)}{2}$

Limit of a sum

February 8, 2019March 17, 2019 Tolaso

Let $a_n$ be a real sequence such that $a_n>0$ , $\liminf a_n=1$ , $\limsup a_n =2$ and $\lim \limits_{n \rightarrow +\infty} \sqrt[n] {\prod \limits_{k=1}^n{a_k}}=1$ . Prove that

$\lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^{n} a_k =1$

Solution

Lemma: Let $a_n$ be a bounded sequence of, say, complex numbers and let $\ell$ be another complex number.

$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n a_k=\ell \Leftrightarrow \text{D-lim } a_n=\ell$

where the latter means “convergence on a set of density $1$ ” i.e. there exists a subsequence $n_1,n_2,\dotsc$ such that $\lim \limits_{k \to +\infty} a_{n_k}=\ell$ and $n_k$ is dense i.e.

$\lim \limits_{n \to +\infty} \frac{\vert \{1,2,\dotsc,n\} \cap \{n_1,n_2,\dotsc,\}\vert}{n}=1$

Proof: The proof is omitted because it is too technical.

Hence,

$\begin{align*} \lim_{n \rightarrow +\infty} \sqrt[n]{\prod_{k=1}^n a_k}=1 &\Leftrightarrow \lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^n \log a_k=0 \\ &\Leftrightarrow \text{D-lim } \log a_n=0 \\ &\Leftrightarrow \text{D-lim } a_n=1 \\ &\Leftrightarrow \lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^n a_k=0 \end{align*}$

All we need is that the logarithm is continuous and that $a_n$ as well as $\log a_n$ are bounded sequences.

Note: We can simplify the conditions to $a_n>0, \liminf \limits_{n \rightarrow +\infty} a_n>0$ and $\limsup \limits_{n \rightarrow +\infty} a_n< +\infty$ .

The exercise along with the solution may be found on AoPS.com . The proof of the claim may also be found there.

Convergent sequence

January 12, 2019 Tolaso

Let $\{x_n\}_{n \in \mathbb{N}}$ be a sequence of positive real numbers such that

(1) $\begin{equation*} x_{n+m} \leq x_n + x_m \quad , \quad m, n \in \mathbb{N} \end{equation*}$

Prove that $\left\{\dfrac{x_n}{n} \right\}_{n \in \mathbb{N}}$ converges.

Solution

Fix $m$ and let $n \geq m$ . Then, there exist $k, r$ such that $n=km+r$ where $0\leq r <m$ . Thus,

$\frac {x_n}{n} = \frac {x_ {km+r}}{n}\leq \frac {kx_ {m}}{n} + \frac {x_{r}}{n}$

Letting $n \rightarrow +\infty$ it follows that

$\limsup \frac {x_n}{n} \leq \frac {x_ {m}}{m} + 0$

Since this holds forall $m$ it follows that $\displaystyle \limsup \frac {x_n}{n} \leq \liminf \frac {x_ {m}}{m}$ and the result follows.