# Möbius Mania

$\sum_{n=1}^{\infty} \mu (n)$

1] Prove/Disprove , that the sum mentioned above converges.

Also ,

$\sum_{n=1}^{\infty}(-1) ^{\mu (n)}$

2] Prove / Disprove , the sum mentioned above converges.

2 years, 5 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

The second sum follows suit.

- 2 years, 5 months ago

EDIT: My new thinking is that both sums does not converge or diverge since it's value fluctuates.

Either ways, here's my approach:

Let $$f$$ be a completely multiplicative function.

$\sum_{n>0}f(n)=\left[\prod_{p \text{ is prime}}(1-f(p))\right]^{-1}$

Through euler product.

Expanding the product gives

$\prod_{p \text{ is prime}}(1-f(p))=\sum_{n>0}\mu(n)f(n)$

Putting it all together

$\sum_{n>0}f(n)=\left[\sum_{n>0}\mu(n)f(n)\right]^{-1}$

Substituting $$f=1$$ gives

$\sum_{n>0}1=\left[\sum_{n>0}\mu(n)\right]^{-1}$

$\sum_{n>0}\mu(n)=0$

Of course there is a lot of hand waving here.

- 2 years, 5 months ago

Diverges means doesnt converge to a specific finite value, so it has to be eother one.

- 2 years, 5 months ago

Ok, so it diverges

- 2 years, 5 months ago

@Julian Poon , @Aareyan Manzoor any modifications ?

- 2 years, 5 months ago

Till some point I also thought like this , but later , I left it as I thought it may be wrong.

- 2 years, 5 months ago

THere is a high possibility my steps aren't justified, since I have assumed the sum converged.

- 2 years, 5 months ago

- 2 years, 5 months ago

Both sums diverges.

- 2 years, 5 months ago

×