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I am interested in knowing if the following series converges or not:

\begin{equation} \sum_{n=1}^{\infty} \prod_{i=1}^n \left(1-e^{-\sqrt{i}} \right) \qquad Expression \; 1 \end{equation}

If that is not well-known, then can someone please tell me something about the converges of

\begin{equation} \sum_{n=1}^{\infty} \prod_{i=1}^n \left(1-e^{-i} \right) \qquad Expression \; 2 \end{equation}

If that is not well-known, then can someone please tell me something about

\begin{equation} \prod_{i=1}^n \left(1-e^{-i} \right) \qquad Expression \; 3 \end{equation}

For instance, is there an analytical expression $F(i)$ which I can relate to expression 3 and, hopefully, work with in determining the converging properties of Expression 2 ?

I already know that expression 3 is a particular case of the q-Pochhammer symbol. I also know it can be related with the Lambert series (whos properties I;m not familiar with). That's basically all I know. Cheers!

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  • $\begingroup$ The sums are divergent as the terms do not tend to zero. If they did, it would imply that one of the $(1-e^i)$ expressions would be zero which is clearly not the case. $\endgroup$ – aleden Feb 12 at 1:48
  • $\begingroup$ Well, the thing is that if you replace $(1-e^{-i})$ by $(1-e^{-a})$ where $a=const>0$, the sums converge (its the geometrical series). However, $(1-e^{-a})$ is never zero, but $(1-e^{-a})^n$ does go to zero as $n$ tends to infinity. Now, in the case which I;m asking $\prod (1-e^{-i})<1$ but I don't know if in the limit this product goes to zero. Thus, @aleden, if you're sure of what you're saying, please elaborate. Thnx $\endgroup$ – Henry Kel Feb 12 at 2:42
  • $\begingroup$ It seems that, as a very first approximation, $\sum_{n=1}^{p} \prod_{i=1}^n \left(1-e^{-i} \right)\sim \frac p 2 $. For $p=10^3$, the summation is $\approx 504.622$ and for $p=10^4$, it is $\approx 5044.48$ $\endgroup$ – Claude Leibovici Feb 12 at 11:39
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I will take this opportunity to share some of the cooler things about elementary symmetric polynomials which seem to apply in this case. Let's forgo the considerations of convergence for the moment, and consider these products symbolically as polynomial expansions in some auxiliary $x$. For your third expression, suppose more generally that we are looking at products of the form $$E_n(x) := \prod_{1 \leq i \leq n} \left(1 - e^{f(i)} x\right).$$ Then we can expand these products for $x_j := e^{f(j)}$ as follows: $$E_n(x) = \sum_{k=0}^n (-1)^k e_k(x_1,x_2,\ldots,x_n) x^k.$$ Now if we define the $k^{th}$ power sum symmetric polynomial by $$p_k(x) := \sum_{i=1}^x x_i^k = \sum_{i=1}^x e^{k \cdot f(i)},$$ we typically can get expansions for the limiting cases of these products as $$\lim_{n \rightarrow \infty} E_n(x) = \exp\left(\sum_{k \geq 1} \frac{(-1)^{k+1}}{k} p_k(\infty) x^k\right),$$ formally at least, and when $x \mapsto 1$ provided suitable convergence conditions on the infinite products.

Now let's examine a slightly more general form of your first two series expressions (again, without immediate considerations of convergence for right now): $$F_N(x) := \sum_{n \leq N} E_n(x) = N + \sum_{k=1}^{N} (N+1-k) (-1)^k e_k \cdot x^k.$$ Thus to evaluate these classes of series, we need to examine the limiting cases of the following equations as $N \rightarrow \infty$: $$\lim_{N \rightarrow \infty} \left[N + (N+1) (E_{N}(1)-1) + [w^N] \left(\sum_{i \geq 1} p_i(\infty) w^i\right) \exp\left(-\sum_{k \geq 1} p_k(\infty) \frac{w^k}{k}\right)\right].$$ This is not a full solution as the corresponding cases of the power sum polynomials, $p_k = \sum_{i \geq 1} e^{k \cdot f(i)}$, are difficult to sum in closed-form, but it should allow you to make some simplifications for special cases of the $f(i)$. For example, I was able to approximate some of these series in the special case of $f(i) := -i$ by making some asymptotic assumptions.

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