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Sum of reciprocal sequence

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:

  1. 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}.

  2. The number 3-\sqrt{5} is known as the Grafting constant.
  3. Under the same assumptions it holds that

        \[\lim_{n \rightarrow +\infty} \left( a_1 a_2 a_3 \dots a_n \right)^{\frac{1}{2^{n}}}=\frac{3+\sqrt{5}}{2}\]

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Limit of a sequence

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,

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

 

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Limit of a sum

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.

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Convergent sequence

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.

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On a nested sin sequence

Consider the sequence x_n defined recursively as

    \[x_1=1 \quad, \quad x_{n+1}=\sin x_n\]

Prove that \lim \limits_{n \rightarrow +\infty} \sqrt{n} x_n = \sqrt{3}.

Solution

Lemma: If a_n is a sequence for which \displaystyle \lim_{n\to+\infty}(a_{n+1}-a_n)=a then

    \[\lim_{n\to + \infty}\frac{a_n}n=a.\]

Proof: In Stolz theorem we set x_{n}=a_{n+1} and y_n=n.

It is easy to see that x_n is is monotonically decreasing to zero. Moreover, an application of L’Hospital’s rule gives

    \[\lim_{x\to 0}\frac{x^2-\sin^2x}{x^2\sin^2x}=\frac{1}{3}\]

Therefore

    \[\lim_{n\to+\infty}\left(\frac1{a_{n+1}^2}-\frac1{a_n^2}\right)=\frac{1}{3}\]

Now, due to the lemma we have \lim\limits_{n\to+\infty} na_n^2 = 3 and the result follows.

Remark : The asymptotic now follows to be \displaystyle x_n \sim \sqrt{\frac{3}{n}}.

Problem: Find what inequality should \beta satisfy such that the series

    \[\mathcal{S}=\sum_{n=1}^{\infty} x_n^\beta\]

converges.

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