Diverges or Converges?

Define $\chi(m) = \begin{cases} +1 & m \equiv 1 \pmod{4} \\ -1 & m \equiv 3 \pmod{4}\end{cases}$.Then, the sum is $\displaystyle\sum_{n = 0}^{\infty} \dfrac{\chi(2n+1)\tau(2n+1)}{2n+1}$.

Notice that $\chi(m), \tau(m), m$ are multiplicative functions, i.e.:

If $m = \displaystyle\prod_{i = 1}^{r}p_i^{e_i}$ for some primes $p_i$, then $\dfrac{\chi(m)\tau(m)}{m} = \displaystyle\prod_{i = 1}^{r}\dfrac{\chi(p_i^{e_i})\tau(p_i^{e_i})}{p_i^{e_i}}$.

So, if we ignore convergence, we can factor the summation in a manner similar to the Euler Product:

$\displaystyle\sum_{m \ \text{is odd}} \dfrac{\chi(m)\tau(m)}{m} = \displaystyle\prod_{\substack{p \ \text{is prime} \\ p \neq 2}} \sum_{k = 0}^{\infty}\dfrac{\chi(p^k)\tau(p^k)}{p^k} = \displaystyle\prod_{\substack{p \ \text{is prime} \\ p \neq 2}} \displaystyle \sum_{k = 0}^{\infty}\dfrac{\chi(p)^k(k+1)}{p^k}$.

$= \displaystyle\prod_{\substack{p \ \text{is prime} \\ p \neq 2}} \dfrac{1}{\left(1-\tfrac{\chi(p)}{p}\right)^2} = \left(\displaystyle\prod_{\substack{p \ \text{is prime} \\ p \neq 2}}\sum_{k = 0}^{\infty} \dfrac{\chi(p)^k}{p^k}\right)^2 = \left(\displaystyle\prod_{\substack{p \ \text{is prime} \\ p \neq 2}}\sum_{k = 0}^{\infty} \dfrac{\chi(p^k)}{p^k}\right)^2$.

Then, by doing the Euler Product technique in reverse, we get:

$\left(\displaystyle\sum_{m \ \text{is odd}} \dfrac{\chi(m)}{m}\right)^2 = \left(\displaystyle\sum_{n = 0}^{\infty} \dfrac{(-1)^n}{2n+1}\right)^2 = \left(\tan^{-1}(1)\right)^2 = \left(\dfrac{\pi}{4}\right)^2 = \boxed{\dfrac{\pi^2}{16}}$.

Also, note that none of this proves convergence.

Thank you Fariz for sharing this question with us. As is probably clear from the number of comments posted for this discussion, this is an extremely interesting question. Not only is it highly nontrivial, but it is also related to several very important ideas in analytic number theory. In particular, Jimmy's first calculation is related to the subject of Dirichlet series (of which Riemann's zeta function is the most famous example). Here I will present an alternative proof of convergence (without computing the actual value of the sum). My primary goal is to illustrate the use of the summation by parts technique, which is incredibly important -- not just in analytic number theory but in many other areas.

To save space, I will be a bit sketchy, but feel free to ask for clarification of any of the steps in the proof.

Step 1. Since the $n$-th term of the series goes to $0$ as $n\to\infty$, it is enough to show that the sequence of the even-numbered partial sums of Fariz's series has a limit, or equivalently, that it is a Cauchy sequence. So let's consider $S_N = \sum_{n=0}^{2N-1} \frac{(-1)^n\,\tau(2n+1)}{2n+1}.$ We can write $S_N = X_N - Y_N$, where $X_N = \sum_{k=0}^{N-1} \frac{\tau(4k+1)}{4k+1}, \ \ \ Y_N = \sum_{k=0}^{N-1} \frac{\tau(4k+3)}{4k+3}.$

Step 2. Let's pause for a second and try to understand why proving convergence is so difficult. The individual sequences $X_N$ and $Y_N$ both tend to $\infty$ as $N\to\infty$, so to show that $S_N$ has a limit, we need to show (informally speaking) that $X_N$ and $Y_N$ grow at approximately the same rate (this will be made completely precise below). The reason it is hard to estimate the rate of growth of $X_N$ is that the divisor function $\tau(k)$ oscillates rather wildly: for instance, it achieves arbitrarily high values, but at the same time, it also hits the value $2$ infinitely often. This is where summation by parts comes to the rescue. It is a very general method for improving the behavior of sums of oscillating terms. (Summation by parts is the discrete analogue of integration by parts, which is even more important, and can likewise be used to handle oscillating integrals.) We will use the following version of summation by parts, which you can easily check. Suppose that $a_k$ and $c_k$ are sequences of real or complex numbers ($k\geq 0)$. Then for all integers $N>M>0$, $\sum_{k=M}^{N-1} a_k c_k = A_N c_{N-1} - A_M c_{M-1} + \sum_{k=M}^{N-1} A_k (c_{k-1} - c_k)$ where $A_k = \sum_{j=0}^{k-1} a_j$ for all $k\geq 1$ (to prove this, write $a_k=A_{k+1}-A_k$).

Step 3. Let us apply the above formula with $a_k=\tau(4k+1)$ and $c_k=\frac{1}{4k+1}$. We have $c_{k-1} - c_k = \frac{1}{4k-3} - \frac{1}{4k+1} = \frac{4}{(4k-3)(4k+1)} = \frac{1}{4}\,k^{-2} + O(k^{-3}),$ so with the notation introduced earlier, we obtain $X_N - X_M = \sum_{k=M}^{N-1} \frac{\tau(4k+1)}{4k+1} = \frac{A_N}{4N-3} - \frac{A_M}{4M-3} + \frac{1}{4}\,\sum_{k=M}^{N-1} A_k \cdot (k^{-2} + O(k^{-3})),$ where $A_k = \sum_{j=0}^{k-1} \tau(4j+1)$. A similar calculation shows that $Y_N - Y_M = \sum_{k=M}^{N-1} \frac{\tau(4k+3)}{4k+3} = \frac{B_N}{4N-1} - \frac{B_M}{4M-1} + \frac{1}{4}\,\sum_{k=M}^{N-1} B_k \cdot (k^{-2} + O(k^{-3})),$ where $B_k = \sum_{j=0}^{k-1} \tau(4j+3)$.

Step 4. Here comes the second crucial ingredient: using the idea of the Dirichlet hyperbola method to estimate the rate of growth of the sequences $A_k$ and $B_k$. This step is very closely related to the approach used in Jimmy's second calculation. I leave it as a very instructive exercise to modify the argument presented here to prove the following estimates: $\sum_{j=0}^{N-1} \tau(4j+1) = \frac{1}{2} N\,\log(N) + \frac{2\gamma + 4\log(2)-1}{2} N + O(\sqrt{N})$ and $\sum_{j=0}^{N-1} \tau(4j+3) = \frac{1}{2} N\,\log(N) + \frac{2\gamma + 4\log(2)-1}{2} N + O(\sqrt{N}),$ where $\gamma$ is Euler's constant.

Step 5. Now we are essentially done. Using the asymptotic formulas for $A_k$ and $B_k$ given above (which imply, in particular, that $A_k-B_k=O(\sqrt{k})$ and that $A_k$ and $B_k$ are both $O(k\,\log(k))$), we obtain, after various straightforward manipulations: $\begin{aligned} S_N - S_M &=& (X_N-X_M) - (Y_N-Y_M) \\ &=& O(N^{-1/2}\,\log(N)) + O(M^{-1/2}\,\log(M)) + \sum_{k=M}^{N-1} O(k^{-3/2}), \end{aligned}$ and this is enough to conclude that $\{S_N\}$ is a Cauchy sequence. EDIT: as pointed out below, I was careless/pessimistic: in fact, we get a better estimate $S_N - S_M = O(N^{-1/2}) + O(M^{-1/2})$.

Here is my attempt: I did not use the Euler Product and I think my solution also explains why the sum converges.

The basic idea looks like this:

Split the divisor function into parts, then sum them up.

Define $\tau_{1}(n) = 1$ if n is divisible by 1, and 0 otherwise. Clearly, this will always evaluate for 1. Also, define $\tau_{3}(n) = 1$ if n is divisible by 3, and 0 otherwise. And so on.

Clearly, $\tau(n) = \sum\limits_{a=1}^ \infty \tau_{a}(n)$.

Note the following, when summing up all the values of $2n+1$ divisible by some $2a+1$, all such positive values of n are generated by $(2a+1)k + a$ for $k \geq 0$

Now we can rewrite the original sum as $\sum\limits _{a=0}^{\infty } \sum\limits _{k=0}^{\infty } \frac{(-1)^{(2 a+1) k+a}}{2 ((2 a+1) k+a)+1}$

Now if we only evaluate $\sum _{k=0}^{\infty } \frac{(-1)^{(2 a+1) k+a}}{2 ((2 a+1) k+a)+1}$ for some fixed a, we will get $\frac{\tan ^{-1}\left((-1)^a\right)}{2 a+1}$.

Note that $\tan ^{-1}\left((-1)^a\right)=\frac{1}{4} \pi (-1)^a$

If we evaluate this for all a, we will get $\pi^2/16$

I am really sorry for poor formatting. Please reply to this if you did not understand what I meant.

Using a computer program, it can be verified that for $50000 \le N \le 100000$ we have

$0.616343 < \displaystyle\sum_{n = 0}^{N} \dfrac{(-1)^n \tau(2n+1)}{2n+1} < 0.617383$.

This suggests that it converges, but is not definitive proof.

The sum does not converge absolutely, which means clever rearrangements could change convergence. Also, the magnitude of the terms is not strictly decreasing, so the alternating series test cannot be applied.

The blog post inspired by this question is now live

On a related note, are there any theorems out there relating divisors to the magnitude of numbers, kind of like how x/ln(x) relates the number of primes? I think that would be a useful information to have to go solve this problem.

Another solution, though it is not very different from the ones provided in the fantastic discussion in this note:

$A=\sum_{n\ge 0}\frac{(-1)^n}{2n+1}\sum_{(2d+1)|2n+1}1\\=\sum_{d\ge 0}\sum_{n:2d+1|2n+1}\frac{(-1)^n}{2n+1}\\=\sum_{d\ge 0}\sum_{l\ge 0}\frac{(-1)^{((2d+1)(2l+1)-1)/2}}{(2d+1)(2l+1)}\\=\sum_{d\ge 0}\sum_{l\ge 0}\frac{(-1)^{d+l}}{(2d+1)(2l+1)}\\=\left(\sum_{d\ge 0}\frac{(-1)^d}{2d+1}\right)^2=\left(\arctan(1)\right)^2=\left(\frac{\pi}{4}\right)^2$ which gives the answer as $\boxed{\frac{\pi^2}{16}}$.

I think it converges... For example, fixing $2n+1 = 3^{k}$

$\displaystyle \sum_{n=0}^\infty \frac{(-1)^n\tau(2n+1)}{2n+1} = \displaystyle \sum_{k=0}^\infty \frac{(-1)^k\tau(3^{k})}{3^{k}} = \displaystyle \sum_{k=0}^\infty \frac{(-1)^k(k+1)}{3^{k}} \rightarrow 0$

It works for every power of prime, but it needs to be generalised of course...

i think that is divergen which use GF and taylor series