Is this sum constant for n?

Question

Hi I can prove that this sum is constant in $n\in \mathbb{N}$. However my proof is very long (a few pages with probability involved). Does anyone see a simple proof. The sum in question is (a q-series of sorts) With $|\lambda|<1$, $\lambda$ real

$$ \sum_{k=0}^{\infty}\frac{\lambda^{k^{2}+kn}}{\prod_{i=1}^{n+k}(1-\lambda^{i})\prod_{i=1}^{k}(1-\lambda^{i})} $$

EDIT: Some people do not agree the sum is constant in $n$, putting it in matlab for 100 terms in our sum it seems to be constant in $n$? or is my matlab coding that bad ?

Code : run this for different n, and say a=$\lambda$=0.5 when you increase j (i.e how many summands) they converge to the same thing for different n

% function for j-summing up products, with parameter n=n, and parameter
% a=lambda
function [S] = sumfunction(n,j,a)

assert(mod(n,1)==0,'n is not a natural nunmber')
assert(n>=0,'n is not a natural nunmber')
assert(mod(j,1)==0,'m is not a natural nunmber')
assert(j>=0,'m is not a natural nunmber')
assert(a<1,'a is not less than 1')


S=0;

for k = 0:j
   x=1-(a).*a.^((0:k+n-1)) ; % creates a vector with elements that we wish to multiply
  y=1-(a).*a.^((0:k-1)) ;  % creates a vector with elements that we wish to multiply

    S=S+(a^(k^(2) + k*n))/(prod(x)*prod(y)) ; 
end
  S;

end

Answer 1

This is not true, for example when $\lambda=0.5$: Denote the summand by $a^n_k>0\ $ for every $n,k\in\mathbb{N}$, then

$$\frac{a^2_k}{a^1_k}=\frac{\lambda ^k}{1-\lambda ^{k+2}}=\frac{1}{\lambda ^{-k}-\lambda ^2}\leq\frac{1}{\lambda ^{-1}-\lambda ^2}<1$$

for every $k\in\mathbb{N}\ $, therefore

$$\sum_{k=1}^\infty a^1_k>\sum_{k=1}^\infty a^2_k$$

Answer 2

Given the function $$ f_n(x) = \sum_{k \geq 0} \frac{x^{k^2+kn}}{\prod_{i=1}^{k} (1-x^i) \prod_{i=1}^{k+n} (1-x^i)} $$ for integer values of $n\geq0$ and real $|x|<1$. Note, that the product $\prod_{i=1}^{0} (1-x^i) = 1$ by definition. Consider the difference $$ \begin{eqnarray} f_{n+1}(x) - f_n(x) & = & \sum_{k \geq 0} \frac{x^{k^2+k(n+1)}}{\prod_{i=1}^{k} (1-x^i) \prod_{i=1}^{k+(n+1)} (1-x^i)}-\sum_{k \geq 0} \frac{x^{k^2+kn}}{\prod_{i=1}^{k} (1-x^i) \prod_{i=1}^{k+n} (1-x^i)}\\ & = &\sum_{k \geq 0} \frac{x^{k^2+kn}}{\prod_{i=1}^{k} (1-x^i) \prod_{i=1}^{k+n+1} (1-x^i)} \left[x^k - (1 - x^{k+n+1}) \right]\\ & = &\sum_{k \geq 0} \frac{x^{k^2+kn}}{\prod_{i=1}^{k} (1-x^i) \prod_{i=1}^{k+n+1} (1-x^i)} \left[x^{2 k + n + 1} - (1-x^k)(1 - x^{k+n+1}) \right]\\ & \overset{*}{=} &\sum_{k \geq 0} \frac{x^{k^2+kn+2k+n+1}}{\prod_{i=1}^{k} (1-x^i) \prod_{i=1}^{k+n+1} (1-x^i)} -\sum_{k \geq 1} \frac{x^{k^2+kn}(1-x^k)(1 - x^{k+n+1})}{\prod_{i=1}^{k} (1-x^i) \prod_{i=1}^{k+n+1} (1-x^i)}\\ & = &\sum_{k \geq 0} \frac{x^{(k+1)^2+(k+1)n}}{\prod_{i=1}^{(k+1)-1} (1-x^i) \prod_{i=1}^{(k+1)+n} (1-x^i)} -\sum_{k \geq 1} \frac{x^{k^2+kn}}{\prod_{i=1}^{k-1} (1-x^i) \prod_{i=1}^{k+n} (1-x^i)}\\ & = &\sum_{l \geq 1} \frac{x^{l^2+ln}}{\prod_{i=1}^{l-1} (1-x^i) \prod_{i=1}^{l+n} (1-x^i)} -\sum_{k \geq 1} \frac{x^{k^2+kn}}{\prod_{i=1}^{k-1} (1-x^i) \prod_{i=1}^{k+n} (1-x^i)}\\& = & 0 \end{eqnarray} $$ Note that in step $*$ the term for $k=0$ in the second summation would be zero due to the factor $(1-x^0)$, and hence that the sum starts with $k=1$.

Hence it follows that $f_0(x)=f_1(x)=\dots=f_n(x)$ for all integer values and provided the function converges.