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) $

• 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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