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Prove that $$\sum_{n=1}^{\infty}\frac{q^n}{1+q^{2n}}=\sum_{n=1}^{\infty}(d_1(n)-d_3(n))q^n$$ where $d_1(n)$ is the number of divisors of $n$ congruent to $1$ modulo $4$ and $d_3(n)$ is the number of divisors of $n$ congruent to $3$ modulo $4$. I tried using the geometric series and interchanging the sum but that did not yield anything useful. I would prefer a algebraic solution rather than a one that depends on partitions but both are welcome.

Thanks!

metamorphy
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Sorfosh
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1 Answers1

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There is a small typo: $d_1(n) $ is the number of divisors congruent to 1 $\pmod{4}$, and similarly for $d_3(n) $.

Let's open the geometric series:

$$ \sum_n \sum_k q^n (-1) ^k q^{2nk} = \sum_{n, k} q^{n(2k+1) } (-1) ^k$$

Now we should group the terms depending on the exponent of $q$. For a given $m$ , which $n, k$ are such that $n(2k+1) =m$ ?

This happens exactly when we have an odd divisor $d=2k+1$ of $m$. The number $n$ will be then determined by $n= m/d$. Now we group the summands depending on the exponent of $q$ as we promised, then we use the substitution $d=2k+1$ :

$$ \sum_m \sum_{n(2k+1)= m } q^{m} (-1) ^k = \sum_m q^m \sum_{d \mid m \text{ odd}} (-1) ^{\frac{d-1}{2}}$$

Let's further separate the odd divisors in two classes. If $d\equiv 1 \pmod{4}$, then the quantity $(d-1) /2$ is even and the sign appearing is $+1$. If $d\equiv 3 \pmod{4}$, then the sign appearing is $-1$. On balance we will have

$$\sum_{ d \mid m \text{ odd}} (-1) ^{\frac{d-1}{2}} = \sum_{ d \mid m \ \ d \equiv 1 } (+1) + \sum_{ d \mid m \ \ d \equiv 3 } (-1) = d_1(m) -d_3(m) $$

Plugging this into the original equation we get

$$\sum_m q^m \sum_{d \mid m \text{ odd}} (-1) ^{\frac{d-1}{2}} = \sum_m (d_1(m) -d_3(m)) q^m$$

As desired.

Andrea Marino
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  • good one ( +1 ) – asgeige Jan 29 '21 at 00:28
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    Note that the coefficient of $q^m,$ multiplied by $4,$ is the total number of all representations $m = u^2 + v^2.$ In Dickson's little 1929 book. – Will Jagy Jan 29 '21 at 00:30
  • @WillJagy that is actually what I am proving :) – Sorfosh Jan 29 '21 at 00:32
  • Cool! Is there a direct way to prove that $(\sum_u q^{u^2} )^2$ equals the LHS of your post? The latter is the function $\sum_n t_n q^n$, where $t_n $ is the number of ways in which $n$ can be written as a sum of squares. This would be a really nice proof of the Dickson result (which I knew with a kind of complicated proof) – Andrea Marino Jan 29 '21 at 00:48
  • @AndreaMarino I showed that the series you mention is equal to the series in the post. So you completed the last piece of my proof! I will study it in a little bit, thank you for that. You can use Jacobi triple product and $1\Psi_1$ Ramanujan summation to prove this. – Sorfosh Jan 29 '21 at 01:34
  • How did you that? I'm curious :) – Andrea Marino Jan 30 '21 at 21:10
  • @AndreaMarino: for an elementary proof see https://math.stackexchange.com/a/737894/72031 – Paramanand Singh Feb 02 '21 at 08:10
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    @AndreaMarino i am in the process of latexing the proof. I can send it your way if you are still interested after I am done. – Sorfosh Feb 02 '21 at 23:42
  • Thanks! I have looked at the proof with the triple product... It's very long!! – Andrea Marino Feb 03 '21 at 10:30
  • @AndreaMarino a lot of basic stuff in here but the core of the arguments are there (its still a draft) https://www.overleaf.com/read/pcvxyvnndfms – Sorfosh Feb 15 '21 at 15:37
  • @AndreaMarino Also could you explain why we can group the terms according to the exponents of $q$? Is that a theorem? I assume it is allowed as the double series converges absolutely. – Sorfosh Feb 15 '21 at 15:40
  • Formale series are not meant to converge; it's just a clever way of listing a sequence – Andrea Marino Feb 15 '21 at 22:04
  • @AndreaMarino for some reason Stack did not notify my of your response. By Reimann's theorems, you cannot rearrange a (single) sum when the series does not converge absolutely as you can get different results. Here we are doing something akin to rearrangement. However, we have two sums so it is throwing me off as to how we can justify it. – Sorfosh Feb 21 '21 at 16:34
  • I still remind you that series are formal here. Suppose you have for example a countable number of formal series $f_1(q), f_2(q) , \ldots, ... $. How it is defined the $k$-th coefficient of $\sum f_i(q) $? That's exactly the sum of $k$-th coefficients. In our cases however, when you swap the sums, you always have a finite number of stuff involved. Things that resemble a sum swap are just a way of writing the k-th coefficient of a sum. – Andrea Marino Feb 21 '21 at 18:22
  • @AndreaMarino I am not trying to be difficult but can we avoid the use of formal series here? the LHS is a function in $q$ (assume |q|<1) I want to show it is equal to the function on the RHS again for $|q|<1$. I do not know much about formal series. When I have a double sum like that I see say $\lim_{n \to \infty} \sum_{0}^{n}\lim_{k_\to \infty}\sum_{0}^{k} a_{n,k}q^{n+k}$ – Sorfosh Feb 21 '21 at 19:50
  • A formal series is just a series in which you don't have to worry about convergence. Even if you don't know much about formal series, just don't worry :) – Andrea Marino Feb 21 '21 at 20:11