Question

Ridiculous Dot Products.

Answer 1

help

Answer 2

is here

Answer 3

Using the law of cosines,
\[\left( \sum_{n=1}^\infty \frac{1}{n} \right)^2 = \left( \sum_{n=1}^\infty 1*1 \right) \left( \sum_{n=1}^\infty \frac{1}{n^2} \right) \cos^2 \theta\]

\[\frac{\left( \sum_{n=1}^\infty \frac{1}{n} \right)^2 }{\left( \sum_{n=1}^\infty 1*1 \right) } = \frac{\pi^2}{6}\cos^2 \theta\]

\[\frac{\left( \sum_{n=1}^\infty \frac{\tau(n)}{n} \right) }{\left( \sum_{n=1}^\infty 1 \right) } \le \frac{\pi^2}{6} \]

$$\lim_{k \to \infty} \frac{\left( \sum_{n=1}^k \frac{\tau(n)}{n} \right) }{\left( \sum_{n=1}^k 1 \right) } \le \frac{\pi^2}{6} $$

Even though this answer is correct, are these steps actually good?

Answer 4

Also, small aside, I used the Dirichlet convolution \(u \star u = \tau\) to convert that squared harmonic series into just the divisor function:
\[\zeta(1)\zeta(1)=\mathrm{D}(\tau;1)\]

Answer 5

woww so advance lol

Answer 6

@RhondaSommer

Answer 7

plz help @RhondaSommer

Answer 8

what do you need @Tennis5518

Answer 9

help with this person here question! @rhondasommer