Multiplicative Dilemma Part 1

Calculus Level 4

n = 1 ϕ ( n ) τ ( n ) μ ( n ) \Large{ \sum_{n=1}^\infty \frac{\phi(n)}{\tau(n)^{\mu(n)}} }

True or False : The above sum converges.

Definitions :

  • ϕ ( n ) \phi(n) is the number of positive integers not greater than n n that are relatively prime to n n (Euler's totient function).
  • τ ( n ) \tau(n) is the number of positive divisors of n n .
  • μ ( n ) \mu(n) is 0 if n n is not squarefree (divisible by a positive square number greater than 1), and otherwise is equal to ( 1 ) Ω ( n ) (-1)^{\Omega(n)} where Ω ( n ) \Omega(n) is the number of prime divisors of n n .

Try Part 2

I don't know True Not enough information False

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1 solution

Otto Bretscher
Feb 22, 2016

This series (whose terms are all positive) diverges on the prime numbers alone since ϕ ( p ) τ ( p ) μ ( p ) = 2 ( p 1 ) \frac{\phi(p)}{\tau(p)^{\mu(p)}}=2(p-1)

Indeed ! Can we do it by taking limit on u n u n 1 \frac{u_{n}}{u_{n-1}} with n = n=\infty where u n u_n and the other are consecutive terms ... @Otto Bretscher

A Former Brilliant Member - 5 years, 3 months ago

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No, I don't think this limit will exist. Looking at the primes is simple and does the job.

Otto Bretscher - 5 years, 3 months ago

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Yes ,,, It's simple approach ...

A Former Brilliant Member - 5 years, 3 months ago

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