Joy of Mathematics

A finite product

April 10, 2023 Tolaso

Evaluate the product

$\Pi = \prod_{k=1}^{n}\frac{(2k)^4+(2k)^2+1}{(2k-1)^4+(2k-1)^2+1}$

Solution

We are making use of the identity

$x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1) = (x^2 + x + 1)((x-1)^2 + (x-1) + 1)$

Hence,

$\begin{align*} \prod_{k=1}^{n}\frac{(2k)^4+(2k)^2+1}{(2k-1)^4+(2k-1)^2+1} &= \prod_{k=1}^{n} \frac{\cancel{\left ( 4k^2- 2k+1 \right )} \left ( 4k^2+2k+1 \right )}{\left ( 4k^2-6k+3 \right ) \cancel{\left ( 4k^2-2k+1 \right )}} \\ &= \frac{\prod_{k=1}^{n} \left ( 4k^2 + 2k+1 \right )}{\prod_{k=1}^{n} \left ( 4k^2-6k+3 \right )} \\ &= \frac{\prod_{k=1}^{n} \left ( 4k^2 + 2k+1 \right )}{\prod_{k=2}^{n} \left ( 4k^2-6k+3 \right )} \\ &= \frac{\prod_{k=1}^{n} \left ( 4k^2 + 2k+1 \right )}{\prod_{k=1}^{n-1} \left ( 4(k+1)^2-6(k+1)+3 \right )} \\ &= \frac{\prod_{k=1}^{n} \left ( 4k^2 + 2k+1 \right )}{\prod_{k=1}^{n-1} \left ( 4k^2 + 2k+1 \right )} \\ &= 4n^2 +2n+1 \end{align*}$

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:

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

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 segments

March 26, 2023 Tolaso

Given the figure below

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evaluate $\mathrm{AB}^2 + \Gamma \Delta^2$ .

Solution

Solution to be migrated by this link.

Generating function

March 24, 2023 Tolaso

Let $\mathcal{H}_n$ be the $n$ – th harmonic number. Prove that

$2 \sum_{n=1}^{\infty} \frac{\mathcal{H}_{n-1}}{n} x^n = \ln^2 (1-x) \quad \text{forall} \quad \left | x \right |< 1$

Solution

Recalling Cauchy’s product we have successively:

$\begin{align*} \ln^2 (1-x) &= \left ( \sum_{n=1}^{\infty} \frac{x^n}{n} \right ) \left ( \sum_{n=1}^{\infty} \frac{x^n}{n} \right ) \\ &= \sum_{n=1}^{\infty} \left ( \sum_{k=1}^{n-1} \frac{1}{k \left ( n-k \right )} \right )x^n \\ &= \sum_{n=1}^{\infty} \left ( \sum_{k=1}^{n-1} \frac{1}{n} \left ( \frac{1}{k} + \frac{1}{n-k} \right ) \right ) x^n \\ &= \sum_{n=1}^{\infty}\frac{1}{n} \left ( \sum_{k=1}^{n-1} \frac{1}{k} + \sum_{k=1}^{n-1} \frac{1}{n-k} \right ) x^n \\ &= \sum_{n=1}^{\infty} \frac{1}{n} \left ( \mathcal{H}_{n-1} + \mathcal{H}_{n-1} \right ) x^n \\ &= 2 \sum_{n=1}^{\infty} \frac{\mathcal{H}_{n-1}}{n} x^n \end{align*}$