Simplifying a quotient

anemone

MHB POTW Director
Staff member
Simplify $$\displaystyle \frac{\Large 1+\frac{1}{2^a}+\frac{1}{3^a}+\frac{1}{4^a}+\cdots}{\Large1-\frac{1}{2^a}+\frac{1}{3^a}-\frac{1}{4^a}+\cdots}$$ where $a>1$ is a real number.

chisigma

Well-known member
Re: Simplying a quotient

Simplify $$\displaystyle \frac{\Large 1+\frac{1}{2^a}+\frac{1}{3^a}+\frac{1}{4^a}+\cdots}{\Large1-\frac{1}{2^a}+\frac{1}{3^a}-\frac{1}{4^a}+\cdots}$$ 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

$\chi$ $\sigma$

anemone

MHB POTW Director
Staff member
Re: Simplying a quotient

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:

$$\displaystyle \frac{\large 1+\frac{1}{2^a}+\frac{1}{3^a}+\frac{1}{4^a}+\cdots}{\Large1-\frac{1}{2^a}+\frac{1}{3^a}-\frac{1}{4^a}+\cdots}=\frac{\large 1+\frac{1}{2^a}+\frac{1}{3^a}+\frac{1}{4^a}+\cdots}{\large\left(1+\frac{1}{3^a}+\frac{1}{5^a}+\cdots \right)-\left( \frac{1}{2^a}+\frac{1}{4^a}+\frac{1}{6^a}+\cdots \right)}$$

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

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

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

$$\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{2^a}{2^a-2}$$

chisigma

Well-known member
Re: Simplying a quotient

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

$\chi$ $\sigma$

anemone

MHB POTW Director
Staff member
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... 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!