Joy of Mathematics

A convergent sequence

April 9, 2023 Tolaso

Let $\left \{ a_n \right \}_{n \in \mathbb{N}}$ be a strictly increasing sequence of positive integers. Prove that the series $\displaystyle \mathcal{S} = \sum_{i=0}^{\infty} \frac{1}{\left [ a_i, a_{i+1} \right ]}$ , where $\left [ \cdot, \cdot \right ]$ denotes the least common multiple, converges.

Solution

We have successively:

$\begin{align*} \sum_{i=0}^{\infty} \frac{1}{\left [ a_i, a_{i+1} \right ]} &= \sum_{i=0}^{\infty} \frac{\left ( a_i,a_{i+1} \right )}{a_i a_{i+1}} \\ &\leq \sum_{i=0}^{\infty} \frac{a_{i+1} - a_i}{a_i a_{i+1}} \\ &= \sum_{i=0}^{\infty} \left ( \frac{1}{a_i} - \frac{1}{a_{i+1}} \right ) \\ &= \frac{1}{a_0} - \lim_{N \rightarrow +\infty} \frac{1}{a_N} \end{align*}$

Fibonacci series

February 3, 2023 Tolaso

Let $\mathcal{F}_n$ denote the $n$ – th Fibonacci number. Prove that:

$\displaystyle \sum_{n=0}^{\infty} \frac{1}{\mathcal{F}_{2n+1} + 1} = \frac{\sqrt{5}}{2}$ .
$\displaystyle \sum_{n=1}^{\infty} \frac{\mathcal{F}_{2n+1}}{\mathcal{F}^2_{2n+1}-1} = 1$ .
$\displaystyle \sum_{n=1}^{\infty} \frac{1}{\mathcal{F}_{2n+1}-1} = \frac{5 - \sqrt{5}}{2}$ .

Solution

Taking partial sums we see that the series telescopes:
$\begin{align*} \sum_{n=0}^N\frac{1}{1+\mathcal{F}_{2n+1}}&= \sum_{n=0}^N\frac{1}{1+\frac{\varphi^{2n+1}+\varphi^{-(2n+1)}}{\sqrt{5}}} \\ &= \sqrt{5} \sum_{n=0}^{N}\frac{\varphi^{2n+1}}{\varphi^{2(2n+1)}+\sqrt{5}\varphi^{2n+1}+1} \\ &=\sqrt{5} \sum_{n=0}^{N}\frac{\varphi^{2n+1}}{(\varphi^{2n+1}+\varphi)\left( \varphi^{2n+1}+\frac{1}{\varphi}\right)}\\ &= \sqrt{5} \sum_{n=0}^{N}\frac{\varphi^{2n+1}}{(\varphi^{2n}+1)\left( \varphi^{2n+2}+1\right)} \\ &= \frac{\varphi\sqrt{5}}{1-\varphi^2}\sum_{n=0}^N\left(\frac{\varphi^{2n}}{1+\varphi^{2n}}-\frac{\varphi^{2n+2}}{1+\varphi^{2n+2}} \right) \\ &=\sqrt{5}\left(\frac{\varphi^{2N+2}}{1+\varphi^{2N+2}} -\frac{1}{2}\right) \end{align*}$
Using Cassini’s identity we note that:
$\begin{align*} \sum_{n=1}^{\infty} \frac{\mathcal{F}_{2n+1}}{\mathcal{F}^2_{2n+1} - 1} &= \sum_{n=1}^{\infty} \frac{\mathcal{F}_{2n+1}}{\mathcal{F}_{2n} \mathcal{F}_{2n+2}} \\ &= \sum_{n=1}^{\infty} \frac{\mathcal{F}_{2n+2} - \mathcal{F}_{2n}}{\mathcal{F}_{2n} \mathcal{F}_{2n+2}} \\ &= \sum_{n=1}^{\infty} \left ( \frac{1}{\mathcal{F}_{2n}} - \frac{1}{\mathcal{F}_{2n+2}} \right )\\ &= \frac{1}{\mathcal{F}_{2}} \\ &= 1 \end{align*}$
Combining the previous results we have successively:
$\begin{align*} 1 &= \sum_{n=1}^{\infty} \frac{\mathcal{F}_{2n+1}}{\mathcal{F}^2_{2n+1} - 1} \\ &= \sum_{n=1}^{\infty} \frac{\mathcal{F}_{2n+1}}{\left ( \mathcal{F}_{2n+1} - 1 \right ) \left ( \mathcal{F}_{2n+1} + 1 \right )} \\ &= \sum_{n=1}^{\infty} \frac{\mathcal{F}_{2n+1} - 1 + 1}{\left ( \mathcal{F}_{2n+1} - 1 \right ) \left ( \mathcal{F}_{2n+1} + 1 \right )} \\ &= \frac{1}{2} \left ( \sum_{n=1}^{\infty} \frac{1}{\mathcal{F}_{2n+1} - 1} + \sum_{n=1}^{\infty} \frac{1}{\mathcal{F}_{2n+1} + 1} \right ) \end{align*}$

The result follows.

A telescopic (!) series

January 18, 2023January 18, 2023 Tolaso

Prove that

$\sum_{n=-\infty}^{\infty} \arctan \frac{\sinh 1}{\cosh 2n} = \frac{\pi}{2}$

Solution

We have successively:

$\begin{align*} \sum_{n=-\infty}^{\infty} \arctan \frac{\sinh 1}{\cosh 2n} &= \arctan \sinh 1 + 2 \sum_{n=1}^{\infty} \arctan \frac{\sinh 1}{\cosh 2n} \\ &= \arctan \frac{e-e^{-1}}{2} + 2 \sum_{n=1}^{\infty}\arctan \frac{e-e^{-1}}{e^{2n} + e^{-2n}} \\ &= \arctan \frac{e - e^{-1}}{1 + e \cdot e^{-1}} + 2 \sum_{n=1}^{\infty} \arctan \frac{e^{-(2n-1) - e^{-(2n+1)}}}{1 + e^{-4n}} \\ &= \arctan \frac{e - e^{-1}}{1 + e \cdot e^{-1}} + 2 \sum_{n=1}^{\infty} \arctan \frac{e^{-(2n-1) - e^{-(2n+1)}}}{1 + e^{-(2n-1)} e^{-(2n+1)}}\\ &= \left ( \arctan e - \arctan e^{-1} \right ) + \\ &\quad \quad \quad + 2 \sum_{n=1}^{\infty} \left ( \arctan e^{-(2n-1)} - \arctan e^{-(2n+1)} \right ) \\ &= \left ( \arctan e - \arctan e^{-1} \right ) + 2 \arctan e^{-1} \\ &= \arctan e + \arctan e^{-1} \\ &= \frac{\pi}{2} \end{align*}$

An exponential series

December 22, 2022December 22, 2022 Tolaso

Prove that

$\sum _{n=0}^{\infty} \frac {2^n}{3^{2^{n-1}} +1} = \frac{\sqrt{3}+1}{2}$

Solution

We recall the product identity

$\prod_{n=0}^{\infty} \left ( 1 + q^{2^n} \right ) = \frac{1}{1-q} \quad , \quad \left | q \right | < 1$

Thus,

$\begin{align*} \log \prod_{n=0}^{\infty} \left ( 1 + q^{2^n} \right ) = \log \frac{1}{1-q} &\Rightarrow \sum_{n=0}^{\infty} \log \left ( 1 + q^{2^n} \right )= -\log \left ( 1- q \right ) \\ &\Rightarrow \frac{\mathrm{d} }{\mathrm{d} q} \sum_{n=0}^{\infty} \log \left ( 1 + q^{2^n} \right ) = \frac{\mathrm{d} }{\mathrm{d} q} \left ( -\log (1-q) \right ) \\ &\Rightarrow \sum_{n=0}^{\infty} \frac{q^{2^n-1} 2^n}{q^{2^n}+1} = \frac{1}{1-q} \\ &\Rightarrow \sum_{n=0}^{\infty} \frac{q^{2^n}2^n}{q^{2^n}+1} = \frac{q}{1-q} \\ &\Rightarrow \sum_{n=0}^{\infty} \frac{2^n}{q^{-2^n} + 1} = \frac{q}{1-q} \\ &\!\!\!\!\!\!\!\!\!\!\overset{q=1/\sqrt{3}}{=\! =\! =\! =\! =\!\Rightarrow } \sum_{n=0}^{\infty} \frac{2^n}{3^{2^{n-1}} + 1} = \frac{\sqrt{3}+1}{2} \end{align*}$

A Möbius inversion sum

November 29, 2022November 29, 2022 Tolaso

Let $q \in \mathbb{R}$ such that $|q|<1$ . Prove that

$\sum_{n=1}^{\infty} \mu(n) \cdot \frac{q^n}{1-q^n} = q$

Solution

We have successively:

$\begin{align*} \sum_{n=1}^{\infty} \mu(n) \cdot \frac{q^n}{1-q^n} &= \sum_{n=1}^{\infty} \mu(n) \sum_{m=0}^{\infty} q^{nm} \\ &= \sum_{n=1}^{\infty} \sum_{d \mid n} \mu(d)q^n \\ &= q + \sum_{n=2}^{\infty} \sum_{d \mid n} \mu(d)q^n \end{align*}$

since $\sum \limits_{d \mid n} \mu(d) =0$ whenever $n \geq 2.$