On the supremum and infimum of a sine sequence

Let \{x_n\}_{n \in \mathbb{N}} be a sequence defined as

    \[x_n = \sin 1 + \sin 3 + \sin 5 + \cdots + \sin (2n -1)\]

Find the supremum as well as the infimum of the sequence x_n.

Solution

Background: This problem was on the shortlist of the 2014 Olimpiada Nationala de Matematica de Romania and was suggested by Leo Giugiuc.

We begin by the very well known manipulation.

    \begin{align*} \sum_{k=1}^n\sin (2k-1) &= {\rm Im}\left[\sum_{k=1}^ne^{i(2k-1)}\right]\\ &= {\rm Im}\left[\frac{e^{i}(1-e^{2ni})}{1-e^{2i}}\right]\\ &= {\rm Im}\left[\frac{e^{i}}{1-e^{2i}}(1-e^{2ni})\right]\\ &= {\rm Im}\left[\frac{e^{i}}{1-e^{2i}}(1-\cos 2n+i\sin 2n)\right]\\ &= {\rm Im}\left[\frac{i}{2\sin 1}(1-\cos 2n+i\sin 2n)\right]\\ &=\frac{1-\cos 2n}{2\sin 1} \end{align*}

Thus \displaystyle x_n= \frac{1-\cos 2n}{2\sin 1} and we have to find the supremum and infimum of \cos 2n. Since the values n \mod 2\pi are dense on the unit circle , the same shall hold for 2 n \mod 2\pi implying that \inf \cos 2n =-1 and \sup \cos 2n = 1. Thus,

\displaystyle \inf\{x_n\}=\frac{1}{2\sin 1}(1-1)=0 \quad , \quad \sup\{x_n\}=\frac{1}{2\sin 1}(1+1)=\frac{1}{\sin 1}

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An arctan series

Evaluate the series

    \[\Omega= \sum_{n=1}^{\infty} \arctan \left ( \frac{9}{9+(3n+5)(3n+8)} \right )\]

(Dan Sitaru)

Solution

Well,

    \begin{align*} \Omega &= \lim_{n\rightarrow\infty}\sum_{k=1}^{n}\arctan\left(\frac{9}{9+(3k+5)(3k+8)}\right)\\ &= \lim_{n\rightarrow\infty}\sum_{k=1}^{n}\left[\arctan\left(k+\frac{8}{3}\right)-\arctan\left(k+\frac{5}{3}\right)\right]\\ &= \lim_{n\rightarrow\infty}\left[\arctan\left(n+\frac{8}{3}\right)-\arctan\left(\frac{8}{3}\right)\right]\\ &=\frac{\pi}{2}-\arctan\frac{8}{3}=\arctan\frac{3}{8}. \end{align*}

and the problem is over.

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Is the infinite union compact?

Let M be a compact metric space and let A_i be a sequence of compact sets in M. Is it true that \bigcup \limits_{i=1}^{\infty} A_i = A is compact?

Solution

The answer is no since we can take A_i = \left[0,1-\frac{1}{i} \right] and M=[0, 1]. Hence

    \[A = \bigcup_{i=1}^{\infty} \left[0,1-\frac{1}{i} \right] = [0, 1)\]

which is clearly not compact.

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A zeta tail limit

Evaluate the limit

    \[\ell = \lim_{n \rightarrow +\infty} n \left( \zeta(2) - \sum_{k=1}^{n} \frac{1}{k^2} \right)\]

Solution

One may use creative telescoping and deduce the double inequality

    \[\frac{1}{n+1}\leq\sum_{k>n}\frac{1}{k^2}\leq \frac{1}{n}\]

since \displaystyle \left(\frac{1}{n}-\frac{1}{n+1}\right)\leq\frac{1}{n^2}\leq \left(\frac{1}{n-1}-\frac{1}{n}\right). The limit follows to be 1.

Another way is using Riemann sum. Note that

    \begin{align*} n \left ( \zeta(2) - \sum_{k=1}^{n} \frac{1}{k^2} \right ) &= n \sum_{k=n+1}^{\infty} \frac{1}{k^2} \\ &= \frac{1}{n} \sum_{k=n+1}^{\infty} \frac{1}{\left ( \frac{k}{n} \right )^2} \\ &\longrightarrow \int_1^{\infty} \frac{{\rm d}x}{x^2}\\ &= 1 \end{align*}

Choose which one you prefer the most.

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A series with least common multiple

Let \{X_n\}_{n \in \mathbb{N}} be a strictly increasing sequence of positive numbers. For all n \geq 1 denote as W_n the least common multiple of the first n terms X_1, X_2, \dots, X_n of the sequence. Prove that , as n \rightarrow +\infty , the following sum converges

    \[\mathcal{S} = \frac{1}{W_1} + \frac{1}{W_2} + \cdots + \frac{1}{W_n}\]

Solution

This is a result due to Paul Erdös stating that if X_1, X_2, \dots, X_n are natural numbers such that 1 \leq X_1 < X_2 < \cdots < X_n then

 \displaystyle \frac{1}{\lcm(X_0,X_1)}+\frac{1}{\lcm(X_1,X_2)}+\cdots+\frac{1}{\lcm(X_{n-1},X_n)}\leq 1-\frac{1}{2^n}

and the original question follows since the sum we seek is less or equal to \displaystyle 1 - \frac{1}{2^n}.

However, we are presenting another proof. Denote as W(n) the average order of the numbers W_n, i.e.,

    \[W (n) = \frac {1} {n} \sum_{k = 1}^{n} W_k.\]

For any k we have W_{k + 1} = W_k \cdot m_k where m_k is the product of primes not present in the factorization of X_1, X_2, \cdots, X_k. Note that m_k are squarefree integers. Note also that it may be an empty product, i.e., m_k = 1. Then

    \[\sum_{k = 1}^{n} W_k = W_1 \cdot \sum_{k = 0}^{n - 1} \prod_{j = 0}^{k} m_j.\]

It is easy to see (and show by induction) that \prod \limits_{j = 0}^{k} m_j > 2^k so we have

    \[\sum_{k = 1}^{n} W_k > W_1 \cdot \sum_{k = 0}^{n - 1} 2^k = (2^n - 1) W_1.\]

Hence, W (n) > \frac {2^n - 1} {n} W_1. Consequently, we have

    \[\sum_{n = 1}^{\infty} \frac {1} {W (n)} < \frac {1} {W_1} \sum_{n = 1}^{\infty} \frac {n} {2^n - 1}\]

So the sum of reciprocals of W (n) converges. Then, by Cesàro summation, we see that

    \[\sum_{n = 1}^{\infty} \frac {1} {W_n}\]

also converges.

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