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I have the following infinite product:

$2/1 * 3/2 * 5/4 * 9/8 * 17/16 * 33/32 * 65/64...$

What does it converge to?

I can take its $\ln()$ to get

$\ln(2) + \ln(3/2) + \ln(5/4)....$

Which using the quotient difference rule resolves to

$S(n) = \ln(2^n + 1) - n\ln(2)$ for $n = 0$ to $\infty$

At this point I don't know how to evaluate this: so I took its derivative to see if I could recognize anything:

$S'(n) = 2^n * ln(2) / (2^n + 1) - ln(2)$

Which really looks like it converges to 0...

Integrating $S(n)$ hasn't been much to avail unless its because I lack knowledge.

What's this come out to?

Amzoti
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Sidharth Ghoshal
  • 16,771

3 Answers3

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Drop the first term for a moment.

$$ \frac{3}{2}\frac{5}{4}\cdots\frac{2^k+1}{2^k}\cdots = \prod_{k=1}^{\infty} (1 + \frac{1}{2^k}) $$

When the product is expanded, what is the coefficient of $\frac{1}{2^n}$? A term comes from any number of distinct factors of the form $\frac1{2^k}$. And so, each partition

$$ n = k_1 + k_2 + \cdots + k_m $$

provides a term in the $\frac{1}{2^n}$. So, the product is:

$$ \sum_{n=0}^{\infty} \frac{q(n)}{2^n} $$

where $q(n)$ is the number of partitions of $n$ into distinct integers. I don't think this has a closed form.

Eric Jablow
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1

There is no nice answer to this. Here is a way to approximate it. $$\sum_{n=0}^{\infty} \ln\left(1+\dfrac1{2^n}\right) = \sum_{n=0}^{\infty} \sum_{k=1}^{\infty}(-1)^{k-1} \dfrac{1}{k2^{kn}} = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}k \sum_{n=0}^{\infty} \dfrac1{2^{kn}} = \sum_{k=1}^{\infty} \left(\dfrac{(-1)^{k-1}}k \dfrac{2^k}{2^k-1} \right)$$

0

4.768462058062755

According to Wolfram Alpha. Look it up it is explained there.

java
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