Solve the limit of the sucession $ \left(1+\frac{1}{2}\right) \left(1+\frac{1}{4}\right) \times \dots \times \left(1+\frac{1}{2^n}\right) $

Question

I need some help with the limit of this succession with general term $$ a_n = \left(1+\frac{1}{2}\right) \left(1+\frac{1}{4}\right) \times \dots \times \left(1+\frac{1}{2^n}\right) $$ Any idea?

Answer 1

As other commentators have pointed out, this product is the q-Pochammer function $(-\tfrac{1}{2}, \tfrac{1}{2})_n$ and so its limit is the value $(-\tfrac{1}{2}, \tfrac{1}{2})_\infty$. Analysis of this type of function will show you that the limit exists, so it remains only to find a good way to compute the limit. In this case it is possible to write out an alternative expression for the logarithm of the limit, as an infinite sum with rapidly decreasing terms. I will show you how to derive this and use it to get a good truncated approximation to the exact limiting value.

First, note that the Maclaurin expansion of the logarithm of a single term is:

$$\ln \Big( 1 + \frac{1}{2^i} \Big) = \sum_{k=1}^\infty (-1)^{k-1} \frac{1}{k} \cdot \frac{1}{2^{i k}}.$$

Thus, taking the logarithm of your product and re-ordering the sums gives the expansion:

$$\begin{equation} \begin{aligned} \ln a_n &= \sum_{i=1}^n \ln \Big( 1 + \frac{1}{2^i} \Big) \\[6pt] &= \sum_{i=1}^n \sum_{k=1}^\infty (-1)^{k-1} \frac{1}{k} \cdot \frac{1}{2^{i k}} \\[6pt] &= \sum_{k=1}^\infty (-1)^{k-1} \frac{1}{k} \sum_{i=1}^n \frac{1}{2^{i k}} \\[6pt] &= \sum_{k=1}^\infty (-1)^{k-1} \frac{1}{k} \cdot \frac{1}{2^k} \frac{1-2^{-n k}}{1-2^{- k}} \\[6pt] &= \sum_{k=1}^\infty (-1)^{k-1} \frac{1}{k} \cdot \frac{1-2^{-n k}}{2^k-1}. \\[6pt] \end{aligned} \end{equation}$$

Taking $n \rightarrow \infty$ then gives the expansion:

$$\ln a_\infty = \sum_{k=1}^\infty (-1)^{k-1} \frac{1}{k} \cdot \frac{1}{2^k-1}. $$

This gives you an exact series form for the limiting value. The terms in this sum shrink rapidly enough to allow you to get a good approximation to the limit by truncating the sum at a reasonably high value. Below I program a function in R to compute the truncated approximation, and then compute this with $10,000$ terms. As you can see, this value matches other computations done in the comment section of your question.

#Create function to compute truncated approximation
A_APPROX <- function(K) {
TERMS <- rep(0, K);
for (k in 1:K) { TERMS[k] <- ((-1)^(k-1))/(k*(2^k-1)); }
exp(sum(TERMS)) }

#Compute truncated approximation with large number of terms
APPROX <- A_APPROX(10000);

#Print truncated approximation to 30 DP
sprintf("%.10f",APPROX);

[1] "2.3842310290"

Answer 2

• You can rewrite it in terms of the Euler function

  $$\psi(x) = \prod_{n=1}^{n=\infty} \left( 1 + x^n\right) = \frac{\phi(x)}{\phi(x^2)}$$

  with $\phi(x) = \prod_{n=1}^{n=\infty} \left( 1 - x^n\right)^{-1}$

• and you can write it as

  $$\psi(x) = \prod_{n=1}^{n=\infty} \left( 1 + x^n\right) = \sum_{n=0}^{n=\infty} b_n x^n$$

  With $b_n$ the number of partitions of $n$ into distinct parts (strict partition). For examples of the partitioning see http://oeis.org/A118457.

See for instance Vaclav Kotesovec A method of finding the asymptotics of q-series based on the convolution of generating functions (but there are many others that describe this)

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the asympote for $b_n$ is calculated to be:

$$b_n \sim \frac{e^{\pi \sqrt{n/3}}}{4 \sqrt[4]{3} {n}^{{3/4}}} \qquad \text{with } n \to \infty$$

see for instance (Ingham 1942 A Tauberian Theorem for Partitions)

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Now we can prove that this sum with the terms $b_n x^n$ at least converges (e.g. ratio test).

Whether it can be calculated exactly I am not sure, but with this summation expression you should be able to compute it and you might also work out a maximum for the error by using an integral.

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