let $|x|<1,|y|>1$, show that $$1+\sum_{n=1}^{\infty}\left((1+x^n)\left(\dfrac{(1-y)(1-yx)(1-yx^2)\cdots(1-yx^{n-1})}{(y-x)(y-x^2)(y-x^3)\cdots(y-x^n)}\right)\right)=0$$

by this sum,I can't it.this problem is from a book,and The author did not give a proof. I fell this reslut is strange,maybe this condition can help me to solve this problem? Thank you

  • $\begingroup$ Hello,I guess this maybe can en.wikipedia.org/wiki/Theta_function $\endgroup$
    – math110
    May 24, 2014 at 6:37
  • $\begingroup$ I don't think it's true for $x=\frac{1}{y}$. $\endgroup$
    – pointer
    May 24, 2014 at 8:12
  • 1
    $\begingroup$ @user121270 "Unfortunately", it is -- if $xy=1$, all terms of the sum are equal to zero, except for the first one whose value turns out to be $(1+x)(1-y)/(y-x)=(1-xy)/(y-x) - 1 = (-1)$. $\endgroup$ May 24, 2014 at 10:16
  • $\begingroup$ I see. I am sorry. $\endgroup$
    – pointer
    May 24, 2014 at 13:41
  • $\begingroup$ I could have sworn I saw this as a past Putnam problem. $\endgroup$
    – anon
    May 24, 2014 at 15:24

2 Answers 2


Let's start by showing the following equality (left-hand side of which is actually the original sum in disguise): $$1+\sum_{n=1}^m \left((1+x^n)\prod_{k=1}^n \frac{1-yx^{k-1}}{y-x^k}\right) = \prod_{k=1}^m \frac{1-yx^k}{y-x^k}$$

The proof by induction is quite simple: Base case ($m=0$) is trivial. For the inductive one, it's sufficient to show the following:

$$ \prod_{k=1}^{m+1} \frac{1-yx^k}{y-x^k} - \prod_{k=1}^m \frac{1-yx^k}{y-x^k} = (1+x^{m+1})\prod_{k=1}^{m+1} \frac{1-yx^{k-1}}{y-x^k}$$

Getting rid of the denominators gives us: $$ \prod_{k=1}^{m+1} (1-yx^k) - (y-x^{m+1})\prod_{k=1}^m (1-yx^k) = (1+x^{m+1})\prod_{k=1}^{m+1} (1-yx^{k-1})$$

Most of the factors in the product cancel out too (factors of the form $(1-yx^k)$ for $1\leq k\leq m$ appear in all of the products): $$(1-yx^{m+1}) - (y-x^{m+1}) = (1+x^{m+1})(1-y)$$

Since the left-hand and right-hand side are equal, we're done with the proof of the equality.

Now that we know the sum is actually equal to a product, let's estimate the terms in the product! Since $|x|<1$, the value of $yx^k$ decreases (in absolute value), so the numerator $(1-yx^k)$ converges to $1$. Similarly, the denominator $(y-x^k)$ converges to $y$, so the whole fraction converges to $\frac{1}{y}$. But that means that after certain point all the terms in the product are, in absolute value, strictly smaller than $\frac{1}{|y|}$. But that means their product is going to converge to zero... and so will the whole product.


The problem appears much harder than it is: in fact, it is susceptible to the most basic technique of writing out a few of the partial sums, guessing their general form, then proving it by induction.

Do this first.

Finally, we need to prove the resulting expression $\to0$ with the hypotheses. Interpret the partial sums as partial products, and show the $n$th factor in the infinite product is asymptotic to $[?]$.


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