Understanding Mathematics: playing with Dirichlet series

Most of these problems are direct results of thinking of numbers in a sieve-like fashion (like Sieve of Eratosthenes). 1. This sieve will go through all multiples $n$ of $a$ and increment the sum by $\frac{1}{n^4}$ for each value of $a$. Let $ab=n$. $\begin{aligned} \sum_{n=1}^\infty\frac{\tau(n)}{n^4} &= \sum_{a=1}^\infty\sum_{b=1}^\infty\frac{1}{(ab)^4} \\ &= \sum_{a=1}^\infty\left[\frac{1}{a^4}\sum_{b=1}^\infty\frac{1}{b^4}\right] \\ &= \left[\sum_{k=1}^\infty\frac{1}{k^2}\right]^2 \\ &= \frac{\pi^8}{8100} \end{aligned}$ 2. This sieve will go through all multiples $n$ of $a$ and increment the sum by $\frac{1}{n}$ if $n$ is odd and decrement the sum by $\frac{1}{n}$ if $n$ is even. $\sum_{n=1}^\infty\frac{(-1)^{n-1}\tau(n)}{n} = \sum_{a=1}^\infty\sum_{b=1}^\infty\frac{(-1)^{ab-1}}{ab}$ Now we know that $ab-1$ will only be even if both $a$ and $b$ are odd. In this case, let $a=2i-1$ and $b=2j-1$ So, we have $\begin{aligned} \sum_{a=1}^\infty\sum_{b=1}^\infty\frac{(-1)^{ab-1}}{ab} &= -\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{1}{ab} + \sum_{i=1}^\infty\sum_{j=1}^\infty\frac{2}{(2i-1)(2j-1)} \\ &= -\left[\sum_{a=1}^\infty\frac{1}{a}\right]^2 + 2\left[\sum_{i=1}^\infty\frac{1}{(2i-1)}\right]^2 \\ &= \left(\sum_{n=1}^\infty\left[-\frac{1}{n}+\frac{\sqrt{2}}{2n-1}\right]\right)\left(\sum_{n=1}^\infty\left[\frac{1}{n}+\frac{\sqrt{2}}{2n-1}\right]\right) \end{aligned}$ We can see that the both factors diverge in magnitude, so this series diverges. 3. As we know $2n+1$ is odd, we will use a sieve similar to that used in (1), but letting $2n+1=(2a-1)(2b-1)$. $\begin{aligned} \sum_{n=1}^\infty\frac{\tau(2n+1)}{(2n+1)^2} &= -1+\sum_{n=1}^\infty\frac{\tau(2n-1)}{(2n-1)^2} \\ &= -1+\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{1}{((2a-1)(2b-1))^2} \\ &= -1+\left[\sum_{a=1}^\infty\frac{1}{(2a-1)^2}\right]^2 \end{aligned}$ To evaluate this summation, note that $\sum_{n=1}^\infty\frac{1}{n^2}=\sum_{n=1}^\infty\left[\frac{1}{(2n-1)^2}+\frac{1}{(2n)^2}\right]$ so $\sum_{n=1}^\infty\frac{\tau(2n+1)}{(2n+1)^2}=-1+\left(\frac{3}{4}\frac{\pi^2}{6}\right)^2=\frac{\pi^4}{64}-1$ 4. Using methods similar to those in (1) and (3), let $a$ and $b$ be integers such that $(2a-1)b=n$. $\sum_{n=1}^\infty\frac{\tau_2(n)}{n^2}=\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{1}{((2a-1)b)^2}=\frac{\pi^4}{48}$ 5. This one is a bit more intuitive than the others. Since each number divides $1$, we know $\tau_2(n)\ge1$, meaning it diverges by comparison: $\sum_{n=1}^\infty\frac{\tau_2(n)}{n}\gt\sum_{n=1}^\infty\frac{1}{n}=\infty$ 6. Think every three numbers: $\frac{1}{3k-2}+\frac{1}{3k-1}-\frac{1}{3k}>\frac{1}{3k}$ By comparison to $\sum_{k=1}^\infty\frac{1}{3k}$, this sum diverges. 7. Similar to the method in (2), let's just find the sum of $\frac{1}{n^2}$ and subtract the multiples of $3$ twice so we get the desired results: $\sum_{n=1}^\infty\frac{1}{n^2}-\sum_{n=1}^\infty\frac{2}{(3n)^2}=\frac{7\pi^2}{54}$ I'm still working on 8, 9, 10, and 11... insights anyone?

John S., Thank you very very much! When I first glanced at the questions, they seems to me to be very difficult. After I read the blog post, I solved them in a few seconds! I'm amazed by analytic number theory, because the result seems to appear in a magic way.

I'm not very practical with analytic calculus; do you think I could read a book about analytic number theory? In that case, could you suggest me one?

I've solved (although not very rigorously) the first seven questions without using Dirichlet series. I only used the facts $\sum _{k=1}^{\infty } \frac{1}{k^2}=\frac{\pi ^2}{6}$ and $\sum _{k=1}^{\infty } \frac{1}{k^4}=\frac{\pi ^4}{90}$

What is Dirichlet Series?

can I make discussion after some months. it will take time for me to study the chapter.

Can somebody help me with Question 2 and 8?