The best way to understand mathematics is to *do* mathematics. In keeping with this principle, this week's discussion offers you a series of questions and problems that further develop the ideas presented in this blog post. If you'd like to discuss your ideas and solutions with your Brilliant fellows, please post them below. I will also comment on them wherever appropriate.

Calculate \(\sum_{n=1}^\infty \frac{\tau(n)}{n^4}\), where \(\tau(n)\) is the number of positive integer divisors of \(n\).

Does the series \(\sum_{n=1}^\infty \frac{(-1)^{n-1}\,\tau(n)}{n}\) converge? If so, what is its sum?

Compute \(\sum_{n=1}^\infty \frac{\tau(2n+1)}{(2n+1)^2}\).

For an integer \(n\geq 1\), let \(\tau_2(n)\) denote the number of all

*odd*positive integer divisors of \(n\). Find the sum \(\sum_{n=1}^\infty \frac{\tau_2(n)}{n^2}\).Does the series \(\sum_{n=1}^\infty \frac{\tau_2(n)}{n}\) converge?

For an integer \(n\geq 1\), let's define \(s(n)=1\) if \(n\) is divisible by \(3\) and \(s(n)=-1\) otherwise. Does the series \(\sum_{n=1}^\infty \frac{s(n)}{n}\) converge? If so, what is the sum?

Same question for the series \(\sum_{n=1}^\infty\frac{s(n)}{n^2}\).

Prove that the series \[ \sum_{n=0}^\infty \frac{(-1)^n \, \tau(2n+1)}{(2n+1)^s} \] converges conditionally (but not absolutely) for any real number \(s>\frac{1}{2}\), and converges absolutely for any \(s>1\).

Investigate the convergence of the last series for \(s\leq \frac{1}{2}\).

Prove the following analogue of Dirichlet's asymptotic formula: as \(N\to\infty\) \[ \sum_{k=0}^{N-1} \tau(2k+1) = \frac{1}{2}\,N\log(N) + \frac{2\gamma+3\log(2)-1}{2}N + O(\sqrt{N}), \] where \(\gamma\) is Euler's constant.

What is the analogue of Dirichlet's asymptotic formula for a sum of the form \( \sum_{k=0}^{N-1} \tau(qk+r) \), where \(0<r<q\) are fixed integers and \(r,q\) are relatively prime?

## Comments

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TopNewestMost 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 \(ab=n\). \[\begin{align*} \sum_{n=1}^\infty\frac{\tau(n)}{n^4} &= \sum_{a=1}^\infty\sum_{b=1}^\infty\frac{1}{(ab)^4} \\ &= \sum_{a=1}^\infty\left[\frac{1}{a^4}\sum_{b=1}^\infty\frac{1}{b^4}\right] \\ &= \left[\sum_{k=1}^\infty\frac{1}{k^2}\right]^2 \\ &= \frac{\pi^8}{8100} \end{align*}\] 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 \[\begin{align*} \sum_{a=1}^\infty\sum_{b=1}^\infty\frac{(-1)^{ab-1}}{ab} &= -\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{1}{ab} + \sum_{i=1}^\infty\sum_{j=1}^\infty\frac{2}{(2i-1)(2j-1)} \\ &= -\left[\sum_{a=1}^\infty\frac{1}{a}\right]^2 + 2\left[\sum_{i=1}^\infty\frac{1}{(2i-1)}\right]^2 \\ &= \left(\sum_{n=1}^\infty\left[-\frac{1}{n}+\frac{\sqrt{2}}{2n-1}\right]\right)\left(\sum_{n=1}^\infty\left[\frac{1}{n}+\frac{\sqrt{2}}{2n-1}\right]\right) \end{align*}\] 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 \(2n+1=(2a-1)(2b-1)\). \[\begin{align*} \sum_{n=1}^\infty\frac{\tau(2n+1)}{(2n+1)^2} &= -1+\sum_{n=1}^\infty\frac{\tau(2n-1)}{(2n-1)^2} \\ &= -1+\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{1}{((2a-1)(2b-1))^2} \\ &= -1+\left[\sum_{a=1}^\infty\frac{1}{(2a-1)^2}\right]^2 \end{align*}\] 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? – Cody Johnson · 3 years, 11 months ago

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– Zi Song Yeoh · 3 years, 10 months ago

I think there's a minor typo in P1. :)Log in to reply

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? – Andrea Marino · 4 years ago

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– John Smith Staff · 4 years ago

I'm glad you enjoyed the post, Andrea! If you are interested in analytic number theory and would like to know more, as a starting point I would recommend Tom Apostol's "Introduction to Analytic Number Theory". You may find that this is exactly what you are looking for, but if you decide that you want something different, feel free to ask again here.Log in to reply

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 \( \sum _{k=1}^{\infty } \frac{1}{k^4}=\frac{\pi ^4}{90} \) – Ivan Stošić · 4 years ago

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aboutDirichlet series! For instance, in my view, the two formulas you mentioned in your comment are calculations of \(\zeta(2)\) and \(\zeta(4)\).Certainly, there is an overlap between problems for which one can use Dirichlet series methods and problems in which other methods can be used. However, Dirichlet series provide a very convenient general framework for thinking about questions 1-9, and after you go through these examples, you will be able to apply the same ideas to a wide variety of other situations. – John Smith Staff · 4 years ago

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– Mansi Shukla · 4 years ago

will any one tell me what is Dirichlet series??pls??i am new to this site.and i just still dont know what to do here or how can i get my solution???Log in to reply

this blog post (you can also click on the link in this comment, it's the same URL). That blog entry explains the main ideas you need to approach the problems listed here. – John Smith Staff · 4 years ago

Welcome to Brilliant! Click the link in the first paragraph that saysLog in to reply

Can somebody help me with Question 2 and 8? – Shiva Chidambaram P · 3 years, 12 months ago

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this proof. For Question 2, you can either adapt the calculation given here, which is the faster method, or prove convergence first (using a method similar to the one you would use for question 8) and then compute the sum itself using the strategy I explained in my blog post. – John Smith Staff · 3 years, 12 months ago

Hint: for Question 8 (the conditional convergence part), it is possible to generalize the strategy used inLog in to reply

can I make discussion after some months. it will take time for me to study the chapter. – Subhadeep Biswas · 4 years ago

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– John Smith Staff · 4 years ago

Of course, you are welcome to add comments or start your own discussion at any time.Log in to reply

What is Dirichlet Series? – Immanuel Panjaitan · 4 years ago

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