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A root limit

May 18, 2019 / Leave a comment

Let $a_i \; , \; i =1, 2, \dots, k$ be positive real numbers such that $a_1\geq a_2\geq \cdots \geq a_k$ . Prove that

$\lim_{n \rightarrow +\infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_k^n} = \max\left \{ a_1, a_2 , \dots, a_k \right \}$

Solution

Without loss of generation , let $a_1 =\max\{ a_1, a_2 , \dots , a_k \}$ . Then,

$\begin{align*} a_1 &=(a_1^n)^{1/n}\\ &\leq (a_1^n+\cdots+a_k^n)^{1/n}\\ &=\sqrt[n]{a_1^n \left (1+\left (\frac{a_2}{a_1} \right )^n + ... + \left (\frac{a_k}{a_1} \right )^n \right )} \\ &\leq \sqrt[n]{{a_1}^n \cdot k} \\ &= a_1\sqrt[n]{k} \\ &=a_1 k^{1/n} \end{align*}$

since $a_i \leq a_1$ forall $i=1,2, \dots,k$ . Thus, by the squeeze theorem it follows that

$\lim_{n \rightarrow +\infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_k^n} =a_1 = \max\left \{ a_1, a_2 , \dots, a_k \right \}$

A limit!

May 18, 2019 / 1 Comment on A limit!

Evaluate the limit:

$\ell = \lim_{m\rightarrow +\infty}\sqrt[m]{m+1}\cdot\sqrt[m^2+m]{\binom{m}{1}\binom{m}{2}\cdots\binom{m}{m}}$

Solution

Let $a_m=\sqrt[m^2+m]{\binom{m}{1}\binom{m}{2}\cdots\binom{m}{m}}$ . Then,

$\log a_m = \log \sqrt[m^2+m]{\binom{m}{1}\binom{m}{2}\cdots\binom{m}{m}} = \frac{1}{m^2+m} \sum_{k=1}^{m} \log \binom{m}{k}$

It follows by Stolz–Cesàro that

$\begin{align*} \lim_{m \rightarrow +\infty} \log a_m &= \lim_{m \rightarrow +\infty} \frac{1}{m^2+m} \sum_{k=1}^{m} \log \binom{m}{k} \\ &=\lim_{m \rightarrow +\infty} \frac{\sum \limits_{k=1}^{m+1} \log \binom{m+1}{k} - \sum \limits_{k=1}^{m} \log \binom{m}{k}}{(m+1)^2 + (m+1) -(m^2+m)} \\ &=\lim_{m \rightarrow +\infty} \frac{1}{2m+2} \sum_{k=1}^{m} \log \frac{m+1}{m+1-k} \\ &= \lim_{m \rightarrow +\infty} \frac{\log \frac{(m+1)^m}{m!}}{2m+2}\\ &=\lim_{m \rightarrow +\infty} \frac{1}{2} \cdot \frac{m}{m+1} \cdot \log \frac{m+1}{\sqrt[m]{m!}} \\ &= \frac{1}{2} \cdot 1 \cdot \log e \\ &= \frac{1}{2} \end{align*}$

In addition,

$\begin{align*} \lim_{m \rightarrow +\infty} \sqrt[m]{m+1} &= \lim_{m \rightarrow +\infty} \left ( m+1 \right )^{1/m} \\ &=\lim_{m \rightarrow +\infty} e^{\frac{\log m}{m+1}} \\ &=1 \end{align*}$

Hence $\ell = \sqrt{e}$ .

Floor series

May 14, 2019 / Leave a comment

Let $\left \lfloor \cdot \right \rfloor$ denote the floor function. Evaluate the series

$\mathcal{S} = \sum_{k=1}^{\infty} \frac{(-1)^{\left \lfloor \sqrt{k} + \sqrt{k+1} \right \rfloor}}{k(k+1)}$

Solution

First of all we note that $4k+3$ and $4k+2$ are never squares. Thus, there exists a positive integer $m$ such that

$m^2 \leq \sqrt{4k+1} < \sqrt{4k+2}< \sqrt{4k+3} < \left ( m+1 \right )^2$

It is easy to see that $\sqrt{4x+1} \leq\sqrt{x} + \sqrt{x+1} < \sqrt{4x+3}$ and thus we conclude that

$\left \lfloor \sqrt{k} + \sqrt{k+1} \right \rfloor = \left \lfloor 4k+1 \right \rfloor$

Now $\left \lfloor 4k+1 \right \rfloor$ is equal to the even number $2n$ if-f

$\left ( 2n \right )^2 \leq 4k+1 < \left ( 2n+1 \right )^2 \Leftrightarrow k \in \left [ n^2, n^2+n \right )$

Hence, since the series is absolutely convergent we can rearrange the terms and by noting that the finite sums are telescopic , we get that:

$\begin{align*} \sum_{k=1}^{\infty} \frac{(-1)^{\left \lfloor \sqrt{k} + \sqrt{k+1} \right \rfloor}}{k(k+1)} &= \sum_{n=1}^{\infty} \bigg ( \sum_{k=n^2}^{n^2+n-1} \left ( \frac{1}{k} - \frac{1}{k+1} \right )- \\ &\quad \quad \quad \quad \quad - \sum_{k=n^2+n}^{\left ( n+1 \right )^2-1}\left ( \frac{1}{k} - \frac{1}{k+1} \right ) \bigg )\\ &=\sum_{n=1}^{\infty} \left ( \frac{1}{n^2} -\frac{1}{n^2+n}-\frac{1}{n^2+n} +\frac{1}{(n+1)^2} \right ) \\ &=\sum_{n=1}^{\infty} \frac{1}{n^2} + \sum_{n=1}^{\infty} \frac{1}{(n+1)^2} - 2 \sum_{n=1}^{\infty} \frac{1}{n^2+n} \\ &=\frac{\pi^2}{6} + \frac{\pi^2}{6}-1 -2 \sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{n+1} \right ) \\ &=\frac{\pi^2}{3} -1-2 \\ &= \frac{\pi^2}{3}-3 \end{align*}$

Double “identical” series

May 12, 2019 / Leave a comment

Compute the series

$\mathcal{S} = \sum_{n=1}^{\infty} \frac{(-1)^n}{2n-1} \sum_{k=n+1}^{\infty} \frac{(-1)^{k+1}}{2k-1}$

Solution

Successively we have:

$\begin{align*} \sum_{n=1}^{\infty} \frac{(-1)^n}{2n-1} \sum_{k=n+1}^{\infty} \frac{(-1)^{k+1}}{2k-1} &= \sum_{n=1}^{\infty} \frac{(-1)^n}{2n-1} \left ( \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{2k-1} + \sum_{k=1}^{n} \frac{(-1)^k}{2k-1} \right )\\ &=\sum_{n=1}^{\infty} \frac{(-1)^n}{2n-1} \left ( \frac{\pi}{4} + \sum_{k=1}^{n} \frac{(-1)^k}{2k-1} \right ) \\ &= -\frac{\pi^2}{16} + \sum_{n=1}^{\infty} \frac{(-1)^n}{2n-1}\sum_{k=1}^{n} \frac{(-1)^k}{2k-1} \\ &= -\frac{\pi^2}{16} + \frac{1}{2} \bigg ( \sum_{n=1}^{\infty} \frac{(-1)^n}{2n-1}\sum_{k=1}^{\infty} \frac{(-1)^k}{2k-1} +\\ & \quad \quad \quad + \sum_{k=1}^{\infty} \frac{1}{\left ( 2k-1 \right )^2} \bigg ) \\ &= \frac{\pi^2}{32} \end{align*}$

On the geometrical view of an integral

April 24, 2019 / Leave a comment

Evaluate the integral

$\mathcal{J} = \int_0^1 \sqrt{4-x^2} \; \mathrm{d}x$

using geometric methods.

Solution

We are working on the following figure

Rendered by QuickLaTeX.com

Thus,

$\begin{align*} \int_{0}^{1} \sqrt{4-x^2} \; \mathrm{d}x &= \left ( \overset{\triangle}{\mathrm{OAB}} \right ) + \left ( \mathrm{A\overset{\frown}{O} \Gamma} \right )\\ &= \frac{\sqrt{3}}{2} + \pi \cdot 2 ^2 \cdot \frac{30}{360} \\ &= \frac{\sqrt{3}}{2} + \frac{4\pi}{12} \\ &= \frac{\pi}{3} + \frac{\sqrt{3}}{2} \end{align*}$

since the red angle is $60^\circ$ due to the triangle $\mathrm{OAB}$ since $\tan \hat{\mathrm{O}} = \sqrt{3}$ ( $\mathrm{OB}=1 \; , \; \mathrm{AB}=\sqrt{3}$ ). Therefore , the green angle is $180^\circ - 90^\circ - 60^\circ = 30^\circ$ . Finally, the area of the circular sector is equal to