MHB Simplifying a Quotient with Real Number $a>1$

AI Thread Summary
The discussion focuses on simplifying the quotient involving the Riemann Zeta function, specifically $$\frac{1+\frac{1}{2^a}+\frac{1}{3^a}+\cdots}{1-\frac{1}{2^a}+\frac{1}{3^a}-\frac{1}{4^a}+\cdots}$$ for real numbers $a > 1$. Participants identify that the numerator simplifies to $\zeta(a)$, while the denominator simplifies to $\zeta(a)(1 - 2^{1-a})$. The conversation also touches on the Riemann Zeta function's definition and its significance in complex analysis. One participant expresses gratitude for the insights shared about the zeta function and its implications, highlighting a shared interest in the topic. Overall, the thread emphasizes the mathematical beauty and complexity of the zeta function in relation to the problem at hand.
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Simplify $$\frac{\Large 1+\frac{1}{2^a}+\frac{1}{3^a}+\frac{1}{4^a}+\cdots}{\Large1-\frac{1}{2^a}+\frac{1}{3^a}-\frac{1}{4^a}+\cdots}$$ where $a>1$ is a real number.
 
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Re: Simplying a quotient

anemone said:
Simplify $$\frac{\Large 1+\frac{1}{2^a}+\frac{1}{3^a}+\frac{1}{4^a}+\cdots}{\Large1-\frac{1}{2^a}+\frac{1}{3^a}-\frac{1}{4^a}+\cdots}$$ where $a>1$ is a real number.

Is...

$\displaystyle 1 + \frac{1}{2^{a}} + \frac{1}{3^{a}} + \frac{1}{4^{a}} + ... = \zeta(a)\ (1)$

... and...

$\displaystyle 1 - \frac{1}{2^{a}} + \frac{1}{3^{a}} - \frac{1}{4^{a}} + ... = \zeta (a) - 2^{1-a}\ \zeta(a) = \zeta(a)\ (1 - 2^{1-a})\ (2)$

... so that...

Kind regards

$\chi$ $\sigma$
 
Re: Simplying a quotient

chisigma said:
Is...

$\displaystyle 1 + \frac{1}{2^{a}} + \frac{1}{3^{a}} + \frac{1}{4^{a}} + ... = \zeta(a)\ (1)$

... and...

$\displaystyle 1 - \frac{1}{2^{a}} + \frac{1}{3^{a}} - \frac{1}{4^{a}} + ... = \zeta (a) - 2^{1-a}\ \zeta(a) = \zeta(a)\ (1 - 2^{1-a})\ (2)$

... so that...

Kind regards

$\chi$ $\sigma$

Hi chisigma,

Thanks for participating! I can tell that this problem seems like an easy one and probably doesn't count as a challenging problem to you and perhaps some other folks on the forum. :(

I will admit that I don't know what you mean by $\zeta(a)$. I solved the problem purely using an algebraic approach and here is my solution:
$$\frac{\large 1+\frac{1}{2^a}+\frac{1}{3^a}+\frac{1}{4^a}+\cdots}{\Large1-\frac{1}{2^a}+\frac{1}{3^a}-\frac{1}{4^a}+\cdots}=\frac{\large 1+\frac{1}{2^a}+\frac{1}{3^a}+\frac{1}{4^a}+\cdots}{\large\left(1+\frac{1}{3^a}+\frac{1}{5^a}+\cdots \right)-\left( \frac{1}{2^a}+\frac{1}{4^a}+\frac{1}{6^a}+\cdots \right)}$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{\large 1+\frac{1}{2^a}+\frac{1}{3^a}+\frac{1}{4^a}+\cdots}{\large\left(1+\frac{1}{2^a}+\frac{1}{3^a}+\cdots \right)-2\left( \frac{1}{2^a}+\frac{1}{4^a}+\frac{1}{6^a}+\cdots \right)}$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{\large 1+\frac{1}{2^a}+\frac{1}{3^a}+\frac{1}{4^a}+\cdots}{\large\left(1+\frac{1}{2^a}+\frac{1}{3^a}+\cdots \right)-\frac{2}{2^a}\left( 1+\frac{1}{2^a}+\frac{1}{3^a}+\cdots \right)}$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{1}{\large1-\frac{2}{2^a}}$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{2^a}{2^a-2}$$
 
Re: Simplying a quotient

anemone said:
Hi chisigma,

Thanks for participating! I can tell that this problem seems like an easy one and probably doesn't count as a challenging problem to you and perhaps some other folks on the forum. :(

I will admit that I don't know what you mean by $\zeta(a)$...

Hi anemone

Your post is very interesting because illustrates a very important question in the field of complex function. The so called 'Riemann Zeta Function' for $\displaystyle \text{Re}\ (s) > 1$ is defined as...

$\displaystyle \zeta (s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}}\ (1)$

A genial intuition of the Swiss mathematician Leonhard Euler was the fact that $\zeta(s)$ exists for every complex s with the only exception of s=1. Now the series in (1) diverges for $\displaystyle \text{Re}\ (s) \le 1$ and at first we don't see a way to write an explicit expression for $\zeta (s)$ in the left portion of the complex plane. A symple way however is to write the (1) as...

$\displaystyle \zeta(s) = 1 - \frac{1}{2^{s}} + \frac{1}{3^{s}} - \frac{1}{4^{s}} + ... +\ 2\ (\frac{1}{2^{s}} + \frac{1}{4^{s}} + \frac{1}{6^{s}} +...) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}} + 2^{1 - s}\ \zeta (s)\ (2)$

... and from (2)...

$\displaystyle \zeta(s) = \frac{1}{1 - 2^{1-s}}\ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}}\ (3)$

The series in (3) converges for $\displaystyle \text{Re}\ (s) > 0$, so that the range of the $\zeta (s)$ has been extended. The German mathematician Bernard Riemann supposed that all the complex zeroes of $\zeta(s)$ lie on the line $\text{Re}\ (s) = \frac{1}{2}$ and the (3) permits to verify that. If Belle demonstrates that the Riemann's hypothesis is true, there is a rich premium of a million of dollars for Her! ;)...

Kind regards

$\chi$ $\sigma$
 
Hi chisigma,

That is so nice of you to teach me the fundamentals of the zeta function and I really appreciate that and find that it is another very beautiful branch of mathematics which interests me and I hope to discover a beautiful way to prove that hypothesis to be true, even though my knowledge of mathematics is very limited as compared to ALL members of MHB. I am only joking, chisigma...:o How could I prove this elegant and lofty hypothesis to be true, based on my shallow knowledge?

But, I am very happy that you have taught me more about the zeta function, my friend!
 
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