Home » Posts tagged 'Real Analysis'

Tag Archives: Real Analysis

Zeta logarithmic series

September 15, 2019 / Leave a comment

Let \zeta denote the zeta function. Prove that

    \[\sum_{k=1}^{\infty} \left( \log\left(1+\frac{1}{k}\right)-\frac{\zeta(k+1)}{k+1} \right)=0\]

Solution

    \begin{align*} \sum_{k=1}^{\infty} \left ( \frac{\zeta(k+1)}{k+1} - \frac{1}{k+1} \right ) &=\sum_{k= 2}^{\infty} \frac{\zeta(k)-1}{k}\\ &=\sum_{k =2}^{\infty} \sum_{n =2}^{\infty} \frac{1}{kn^k}\\ &=\sum_{n=2}^{\infty} \sum_{k= 2}^{\infty} \frac{1}{kn^k}\\ &=-\sum_{n=2}^{\infty} \left(\ln\left(1-\frac{1}{n}\right)+\frac{1}{n}\right)\\ &=-\sum_{n= 1}^{\infty} \left(\ln\left(1-\frac{1}{n+1}\right)+\frac{1}{n+1}\right) \\ &-\sum_{n =1}^{\infty} \left(\ln\left(\frac{n}{n+1}\right)+\frac{1}{n+1}\right)\\ &=\sum_{n= 1}^{\infty} \left(\ln\left(\frac{n+1}{n}\right)-\frac{1}{n+1}\right) \\ &=\sum_{n= 1}^{\infty} \left(\ln\left(1+\frac{1}{n}\right)-\frac{1}{n+1}\right) \end{align*}

Read more

Mellin transform integral

September 14, 2019 / Leave a comment

Evaluate the integral:

    \[\mathcal{J} = \int_0^\infty x \log x e^{-\sqrt{x}} \, \mathrm{d}x\]

Solution

We are evaluating the Mellin transform of the function f(x)=e^{-\sqrt{x}}.

    \begin{align*} \mathcal{M}\left ( f \right ) &= \int_{0}^{\infty} x^{s-1} e^{-\sqrt{x}} \, \mathrm{d}x \\ &\!\!\!\!\!\overset{u=\sqrt{x}}{=\! =\! =\! =\! =\!} 2\int_{0}^{\infty} u^{2s-1} e^{-u} \, \mathrm{d}u\\ &= 2 \Gamma\left ( 2s \right ) \end{align*}

where \Gamma is the Euler’s Gamma function. Hence,

    \begin{align*} \int_{0}^{\infty} x \log x e^{-\sqrt{x}} \, \mathrm{d}x &= \mathcal{M}'(f) \bigg|_{s=2} \\ &= \left ( 2 \Gamma(2s) \right )'\bigg|_{s=2}\\ &= 4 \Gamma(2s) \psi^{(0)}(2s) \bigg|_{s=2}\\ &=4 \Gamma(4) \psi^{(0)}(4) \\ &= 4 \cdot 6 \cdot \left ( \frac{11}{6} - \gamma \right ) \\ &=44 - 24 \gamma \end{align*}

where \gamma is the Euler – Mascheroni constant and \psi^{(0)} is the digamma.

Read more

Theta Jacobi limit

September 2, 2019 / Leave a comment

Prove that

    \[\lim_{x \rightarrow 1-} \sqrt{1-x} \sum_{n=0}^{\infty} x^{n^2} = \frac{\sqrt{\pi}}{2}\]

Solution

We are working near 1. It follows from the Integral Comparison Test that

    \[\int_0^\infty x^{t^2}\,\mathrm{d}t \leq \sum_{n=0}^\infty x^{n^2} \leq 1 + \int_0^\infty x^{t^2}\,\mathrm{d}t\]

However,

    \begin{align*} \int_0^\infty x^{t^2}\,\mathrm{d}t &= \int_0^\infty \exp\left[-\left(t\sqrt{-\log x}\right)^2\right]\,\mathrm{d}t \\ &= \frac{1}{\sqrt{-\log x}} \int_0^\infty e^{-u^2}\,\mathrm{d}u \\ &= \frac{1}{2} \sqrt{\frac{\pi}{-\log x}} \end{align*}

The result follows from the Sandwich Theorem.

Read more

Double binomial sum

September 2, 2019 / Leave a comment

Evaluate the double sum

    \[\mathcal{S} = \sum_{i=0}^m \sum_{j=0}^n \frac{\binom{m}{i}\binom{n}{j}}{\binom{m+n}{i+j}}\]

Solution

We sum diagonally , hence:

    \begin{align*} \sum_{i=0}^{m}\sum_{j=0}^{n}\frac{\binom{m}{i}\binom{n}{j}}{\binom{m+n}{i+j}}&\overset{i+j=N}{=\! =\! =\! =\!}\sum_{N=0}^{m+n}\sum_{i=0}^m\frac{\binom{m}{i}\binom{n}{N-i}}{\binom{m+n}{N}}\\ &\stackrel{(1)}{=}\sum_{N=0}^{m+n}\sum_{i=0}^{N}\frac{\binom{m}{i}\binom{n}{N-i}}{\binom{m+n}{N}}\\ &=\sum_{N=0}^{m+n}\binom{m+n}{N}^{-1}\sum_{i=0}^{N}\binom{m}{i}\binom{n}{N-i}\\ &\stackrel{(2)}{=} \sum_{N=0}^{m+n}\binom{m+n}{N}^{-1}\binom{m+n}{N}\\ &=m+n+1 \end{align*}

(1): For i=m+1,\ldots,N it holds \binom{m}{i}=0.

(2): \displaystyle \sum_{i=0}^{N}\binom{m}{i}\binom{n}{N-i}=[x^N](1+x)^{m+n}=\binom{m+n}{N}.

Conjecture: Does the following equality

    \[\sum_{a_1=0}^{b_1}\sum_{a_2=0}^{b_2}\cdots\sum_{a_n=0}^{b_n}\frac{\binom{b_1}{a_1}\binom{b_2}{a_2}\cdots\binom{b_n}{a_n}}{\binom{b_1+b_2+\cdots+b_n}{a_1+a_2+\cdots+a_n}}=b_1+b_2+\cdots+b_n+1\]

hold?

Read more

Arctan series

September 2, 2019 / Leave a comment

Evaluate the series

    \[\mathcal{S} = \sum_{n=1}^{\infty} \arctan \frac{1}{n^2}\]

Solution

One can argue that the same technique used to evaluate the sum here can be used here as well. Unfortunately, this is not the case as the sum does not telescope. However the technique used here is the way to go.

First of all we note that

    \[\sum_{n=1}^{\infty} \arctan \frac{1}{n^2} = -\arg \prod _{n=1}^{\infty} \left ( 1 - \frac {i} {n^2}\right )\]

However, it is known that

(1)   \begin{equation*} \sin \pi z =\pi z \prod _{n=1}^{\infty} \left ( 1 - \frac {z^2} {n^2} \right ) \end{equation*}

Let z= \frac {\sqrt {2}}{2} + \frac {i\sqrt {2}}{2}. We note that z^2=i. Hence,

    \begin{align*} \sum_{n=1}^{\infty} \arctan \frac{1}{n^2} &= -\arg \prod _{n=1}^{\infty} \left ( 1 - \frac {i} {n^2}\right ) \\ &=-\arg \left (\frac {\sin \left ( \frac { \pi \sqrt {2}}{2} + \frac {i \pi \sqrt {2}}{2}\right ) } { \frac {\pi \sqrt {2}}{2} + \frac {i \pi\sqrt {2}}{2}} \right ) \end{align*}

which is now a matter of calculations. The sum is equal to

    \[\sum_{n=1}^{\infty} \arctan \frac{1}{n^2} = \frac{\pi }{4} -\arctan \left ( \cot \frac {\pi \sqrt 2}{2} \tanh \frac {\pi \sqrt 2}{2}\right )\]

Read more

Post navigation

← Older posts

Who is Tolaso?

Find out more at his Encyclopedia Page.

Donate to Tolaso Network