# Diverges or Converges?

My friend asked me this interesting problem, but I have no idea how to solve it. Given summation as below:

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

where $\tau(N)$ denotes the number of positive integer divisors of $N$ (you can read here). Determine whether the sum diverges or converges (and if so to what)?

Note by Fariz Azmi Pratama
6 years, 3 months ago

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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.

- 6 years, 3 months ago

Indeed, this is a beautiful calculation! I will add one small comment: if one can show first that the original series converges, then one can actually justify all the steps in your computation, and prove that the final answer is correct. (This is not obvious, since you have to manipulate conditionally convergent series at several steps.) The background for these ideas will be explained in the next Understanding Mathematics discussion, which will be devoted to the Zeta function and Dirichlet series.

Staff - 6 years, 3 months ago

Hello John, is that mean you consider the sum converges so it can produce $\frac{\pi^2}{16}?$

- 6 years, 3 months ago

I'm not sure I understand your question, could you clarify? What I meant by my comment is this. As Jimmy himself pointed out, he did not consider the issue of convergence in his calculation. Because he had to do things like multiplying and rearranging infinite series at several steps of his calculation, this means that in principle, each of these steps has to be rigorously justified, and it is not clear how to do so.

However, it is possible to prove rigorously that if the series you gave us converges at all, then its sum must be equal to $\pi^2/16$. So instead of justifying several manipulations with infinite series, it is enough to prove one convergence statement.

Staff - 6 years, 3 months ago

Oops sorry, I definetely misunderstood your comment. Thanks for your explanation. I'm sure someone here will prove whether the sum converges or diverges. I'm looking forward to it.

- 6 years, 3 months ago

Just figured it out. See edits to above post.

- 6 years, 3 months ago

Beautiful! Thanks Jimmy, you almost answered my question. I think it is converges but still I haven't figured any proper proof yet.

- 6 years, 3 months ago

Jimmy, may I know why you define that by $\pmod{4}?$

- 6 years, 3 months ago

If $2n+1 \equiv 1 \pmod{4}$ then $n \equiv 0 \pmod{2}$ and $(-1)^n = +1$.

If $2n+1 \equiv 3 \pmod{4}$ then $n \equiv 1 \pmod{2}$ and $(-1)^n = -1$.

That is why $\pmod{4}$ is useful there.

- 6 years, 3 months ago

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 $S_N - S_M = O(N^{-1/2}) + O(M^{-1/2})$.

Staff - 6 years, 3 months ago

Hi. Your proof is very good. But, in the last step, when I tried to calculate $S_{N}-S_{M}$, I get it as $O(N^{-\frac{1}{2}})$ and not $O(N^{-\frac{1}{2}}log(N))$. Can you please clarify this?

- 6 years, 3 months ago

EDITED Yes, I think that your calculation is correct, and I inserted the appropriate remark into my proof. Thank you for pointing this out!

Staff - 6 years, 3 months ago

Many thanks, John. It took me a while to undestand about Dirichlet since I'm new in it.

- 6 years, 3 months ago

You're welcome! Feel free to ask more in the discussions (about Dirichlet series or anything else). We are here to show you cool new things in math.

Staff - 6 years, 3 months ago

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.

- 6 years, 3 months ago

Hello, Ivan. Thanks for your great attempt. I think your proof is easy to understand and also well written. But could you explain how you get $\frac{1}{4}\pi(-1)^a = \frac{\pi^2}{16}?$ I'm lost here. Thanks.

- 6 years, 3 months ago

Ok, I will explain:

The original sum is equal to:

$\sum _{a=0}^{\infty } \sum _{k=0}^{\infty } \frac{(-1)^{(2 a+1) k+a}}{2 ((2 a+1) k+a)+1}$

The inner sum can be evaluated to

$\sum _{k=0}^{\infty } \frac{(-1)^{(2 a+1) k+a}}{2 ((2 a+1) k+a)+1}=\frac{\tan ^{-1}\left((-1)^a\right)}{2 a+1}$

We can replace $\tan ^{-1}\left((-1)^a\right)=\frac{1}{4} \pi (-1)^a$ and pull $\frac{\pi }{4}$ out of the sum. If we do, we will get $\frac{1}{4} \pi \sum _{a=0}^{\infty } \frac{(-1)^a}{2 a+1}$ which evaluates to $\frac{\pi ^2}{16}$

Hope this helps

- 6 years, 3 months ago

Definetely helped, thanks.

- 6 years, 3 months ago

My first attempt was something similar to what you did. I managed to not get anywhere then, but now it seems to work out a bit better.

First, let's work with the sum from $n = 0$ to $N$ and then take the limit as $N \to \infty$.

Define $I(k,n) = 1$ if $2k+1 \mid 2n+1$ and $0$ otherwise. Then:

$S_N = \displaystyle\sum_{n = 0}^{N} \dfrac{(-1)^n \tau(2n+1)}{2n+1}$ $= \displaystyle\sum_{n = 0}^{N} \sum_{k = 0}^{n}\dfrac{(-1)^nI(k,n)}{2n+1}$ $= \displaystyle\sum_{k = 0}^{N} \sum_{n = k}^{N}\dfrac{(-1)^nI(k,n)}{2n+1}$

Since the sum is finite, changing the order is valid.

Note that, $2k+1 \mid 2n+1$ iff $2n+1 = (2k+1)(2m+1)$ iff $n = 2km+k+m$ where $m \in \mathbb{N}_0$.

Also, $k \le 2km+k+m \le N$ holds iff $0 \le m \le \dfrac{N-k}{2k+1}$. This gives us:

$\displaystyle\sum_{k = 0}^{N} \sum_{m = 0}^{\left\lfloor \tfrac{N-k}{2k+1}\right\rfloor }\dfrac{(-1)^{2km+k+m}}{(2k+1)(2m+1)}$ $= \displaystyle\sum_{k = 0}^{N} \sum_{m = 0}^{\left\lfloor \tfrac{N-k}{2k+1}\right\rfloor }\dfrac{(-1)^{k+m}}{(2k+1)(2m+1)}$ $= \displaystyle\sum_{k = 0}^{N} \dfrac{(-1)^k}{2k+1}\sum_{m = 0}^{\left\lfloor \tfrac{N-k}{2k+1}\right\rfloor }\dfrac{(-1)^m}{2m+1}$.

Now, we must show that $S_N \to \displaystyle\sum_{k = 0}^{\infty} \dfrac{(-1)^k}{2k+1} \sum_{m = 0}^{\infty} \dfrac{(-1)^m}{2m+1} = \left(\dfrac{\pi}{4}\right)^2 = \dfrac{\pi^2}{16}$ as $N \to \infty$.

However, actually showing this rigorously might not be as easy as it looks.

- 6 years, 3 months ago

So, it turns out that showing that $S_N \to \dfrac{\pi^2}{16}$ as $N \to \infty$ is doable, but very annoying.

The sum $S_N$ is over $X = \left\{(k,m) \in \mathbb{N}^2_0 \ | \ (2k+1)(2m+1) \le 2N+1\right\}$.

Let $A = \left\{(k,m) \in X \ | \ 2k+1 \le \sqrt{2N+1}\right\}$.

Let $B = \left\{(k,m) \in X \ | \ 2m+1 \le \sqrt{2N+1}\right\}$.

Clearly, $X = A \cup B$.

Note that $\displaystyle\sum_{k = 0}^{n} \dfrac{(-1)^k}{2k+1} = \dfrac{\pi}{4} + O\left(\dfrac{1}{n}\right)$ and $\displaystyle\sum_{k = 0}^{n} \dfrac{1}{2k+1} = O(\ln n)$.

Let $L = \left\lfloor \tfrac{1}{2}(\sqrt{2N+1}-1) \right\rfloor$. Note that if $k \le L$ then $\dfrac{N-k}{2k+1} \ge L$. Also, $L = O(\sqrt{N})$.

Then, the sum over $A$ is:

$\displaystyle \sum_{(k,m) \in A} \dfrac{(-1)^{k+m}}{(2k+1)(2m+1)}$ $= \displaystyle\sum_{k = 0}^{L}\left[\dfrac{(-1)^k}{2k+1}\sum_{m = 0}^{\tfrac{N-k}{2k+1}}\dfrac{(-1)^m}{2m+1}\right]$

$= \displaystyle\sum_{k = 0}^{L}\left[\dfrac{(-1)^k}{2k+1}\left(\dfrac{\pi}{4} + O\left(\dfrac{1}{L}\right)\right)\right]$ $= \dfrac{\pi}{4}\displaystyle\sum_{k = 0}^{L}\dfrac{(-1)^k}{2k+1} + O(\ln L)O\left(\dfrac{1}{L}\right)$

$= \dfrac{\pi}{4}\left(\dfrac{\pi}{4} + O\left(\dfrac{1}{L}\right)\right) + O\left(\dfrac{\ln L}{L}\right)$ $= \dfrac{\pi^2}{16} + O\left(\dfrac{\ln L}{L}\right)$ $= \dfrac{\pi^2}{16} + O\left(\dfrac{\ln N}{\sqrt{N}}\right)$.

The same can be done for the sum over $B$ by switching the roles of $k$ and $m$.

The sum over $A \cap B$ is:

$\displaystyle \sum_{(k,m) \in A \cap B} \dfrac{(-1)^{k+m}}{(2k+1)(2m+1)}$ $= \displaystyle\sum_{k = 0}^{L}\left[\dfrac{(-1)^k}{2k+1}\sum_{m = 0}^{L}\dfrac{(-1)^m}{2m+1}\right]$

$= \left[\displaystyle\sum_{k = 0}^{L}\dfrac{(-1)^k}{2k+1}\right]\left[\displaystyle\sum_{m = 0}^{L}\dfrac{(-1)^m}{2m+1}\right]$ $= \left(\dfrac{\pi}{4} + O\left(\dfrac{1}{L}\right)\right)\left(\dfrac{\pi}{4} + O\left(\dfrac{1}{L}\right)\right)$

$= \dfrac{\pi^2}{16} + O\left(\dfrac{1}{L}\right)$ $= \dfrac{\pi^2}{16} + O\left(\dfrac{1}{\sqrt{N}}\right)$.

By Principle of Inclusion-Exclusion, the sum over $X$ is the sum over $A$ plus the sum over $B$ minus the sum over $A \cap B$ which is:

$S_N = 2\left(\dfrac{\pi^2}{16} + O\left(\dfrac{\ln N}{\sqrt{N}}\right)\right) - \left(\dfrac{\pi^2}{16} + O\left(\dfrac{1}{\sqrt{N}}\right)\right)$. $= \dfrac{\pi^2}{16} + O\left(\dfrac{\ln N}{\sqrt{N}}\right)$.

As $N \to \infty$, we have $\dfrac{\ln N}{\sqrt{N}} \to 0$, and thus, $S_N \to \dfrac{\pi^2}{16}$ as desired.

Note that I used Big-O notation here because actually computing an expression for the error terms isn't nearly as important as showing that the error terms tend to zero.

If anyone else can come up with a simpler solution that is rigorous, please post it.

- 6 years, 3 months ago

Superb! Thank you Jimmy, for your amazing solution. I don't know how about the other but I, personally can't prove it any better. Well, I think this problem officially closed :)

- 6 years, 3 months ago

This is a very clever calculation. I am soon going to post another proof of convergence. It is actually a bit longer than your argument (though it shares the same underlying idea), and moreover, unlike your approach, it only shows convergence without computing the value of the sum. So why bother? The method I am going to explain makes it a bit easier to understand what are the key mathematical principles that make it work. Moreover, these principles themselves can be used in a variety of situations, some of which are very different from the one we considered here.

Staff - 6 years, 3 months ago

Wow, I'm waiting for it.

- 6 years, 3 months ago

Wow, very nice. Thanks.

- 6 years, 3 months ago

Could you explain in more detail how you get the conclusion in the line that begins with "Now we can rewrite the original sum as..."? In particular, it seems to me (at least at first glance) that somewhere in this step you are interchanging the order of two infinite summations. If so, how do you justify this step?

Staff - 6 years, 3 months ago

Let's split the tau function into parts, as described.

So, for the first possible divisor, one, we simply evaluate $\sum _{k=0}^{\infty } \frac{(-1)^k}{2 k+1}$

For the second one, three, we shall only take values of 2n+1 divisible by 3. So, our sum will look like this: $\sum _{k=0}^{\infty } \frac{(-1)^{3 k+1}}{2 (3 k+1)+1}$

For the third one, five, the sum will look like this: $\sum _{k=0}^{\infty } \frac{(-1)^{5 k+2}}{2 (5 k+2)+1}$

And so on.

To get to the original sum, we simply sum up all the sums we had here.

And yes, it is possible that I have exchanged the order of summation. I am pretty sure that there is a reason why this should be considered correct, although I am not trained enough in calculus to know why. Maybe I am wrong, who knows?

I just wanted to give a simpler idea not relying on the Euler Product.

P.S.

There is no need to exchange the order of summation. One can keep a in the inside sum and still get the same result. Note that the denominator is equal to (2k+1)(2a+1). So now I am pretty confident that my solution is ok

- 6 years, 3 months ago

Sorry, I don't see this. If you have a proof that does not require interchanging the order of summation, please post the details. On the other hand, in a situation such as this one, where you are dealing with infinite series that converge only conditionally and not absolutely, it is known that interchanging the order of summation is not permissible in general.

Staff - 6 years, 3 months ago

I don't see he interchanging the order of summation, if he does, could you tell me which part of it? Thanks

- 6 years, 3 months ago

If I understand Ivan's approach correctly, it involves replacing $\tau(2n+1)$ in your original series with $\sum_{a=1}^\infty\tau_a(2n+1)$. This is of course perfectly legitimate, but notice that after you do this, the sum over $n$ is the outer sum and the sum over $a$ is the inner sum. However, at some point in the calculation, you can see the different order of summation, where the sum over $a$ became the outer sum.

Staff - 6 years, 3 months ago

Oh didn't notice that, thanks.

- 6 years, 3 months ago

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.

- 6 years, 3 months ago

Hello Jimmy, my friend said that the answer rebound at $\frac{\pi^2}{16}$ which almost same as yours. But I wonder how to solve this problem mathematically, without using algorithm. Any idea?

- 6 years, 3 months ago

When an infinite sum converges to something involving $\pi^2$, the solution often (not always) involves the fact that $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6}$. I'm not sure how to relate that to this problem.

- 6 years, 3 months ago

The blog post inspired by this question is now live

Staff - 6 years, 3 months ago

Wow cool. Nice post!

- 6 years, 3 months ago

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.

- 6 years, 3 months ago

Michael, is this what you are looking for?

- 6 years, 3 months ago

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 $\boxed{\frac{\pi^2}{16}}$.

- 3 years ago

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...

- 6 years, 3 months ago

The value of the sum you computed does not appear to be correct. Consider the power series $\frac{1}{1-x} = \sum_{k=0}^\infty x^k, \quad |x| < 1.$ Then taking derivatives of both sides, we obtain $\frac{1}{(1-x)^2} = \sum_{k=1}^\infty kx^{k-1} = \sum_{k=0}^\infty (k+1)x^k,$ with the same radius of convergence. Then with the choice $x = -1/3$ we obtain $\sum_{k=0}^\infty \frac{(-1)^k (k+1)}{3^k} = \frac{1}{(1+1/3)^2} = \frac{9}{16} > 0.$ Therefore, we cannot infer from your analysis that the original sum $\sum_{n=0}^\infty \frac{(-1)^n \tau(2n+1)}{2n+1}$ is convergent--in fact, your analysis suggests that the answer could be the opposite.

- 6 years, 3 months ago

Yes, checking only for certain value $x$ doesn't mean we can determine whether the sum converges or diverges.

- 6 years, 3 months ago

Using the same logic, one could incorrectly conclude that $\displaystyle\sum_{n = 0}^{\infty}\dfrac{1}{2n+1}$ converges.

- 6 years, 3 months ago

i think that is divergen which use GF and taylor series

- 6 years, 3 months ago

Would you mind to show your solution, Kak Uzu? (Soal Adhel loh hehe ^^)

- 6 years, 3 months ago