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# Summation and Limit?

I got this nice question from my friend.

If

$A = \displaystyle \sum_{n=1}^{\infty} \frac{1}{n(2^n - 1)}$

and

$B_{k} = \frac{2}{1} \times \frac{4}{3} \times \frac{8}{7} \times \ldots \times \frac{2^k}{2^k - 1}$

then prove that

$\lim_{k \to \infty} B_{k} = e^A$

Note by Fariz Azmi Pratama
4 years ago

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Let $$B = \displaystyle\lim_{k \to \infty}B_k$$ $$= \displaystyle\prod_{m = 1}^{\infty} \dfrac{2^m}{2^m-1}$$.

Then, $$\ln B = \displaystyle\sum_{m = 1}^{\infty}\ln\left(\dfrac{2^m}{2^m-1}\right)$$ $$= \displaystyle\sum_{m = 1}^{\infty}-\ln\left(\dfrac{2^m-1}{2^m}\right)$$ $$= \displaystyle\sum_{m = 1}^{\infty}-\ln\left(1 - 2^{-m}\right)$$

$$= \displaystyle\sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty}\dfrac{(2^{-m})^n}{n}$$ $$= \displaystyle\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}\dfrac{(2^{-n})^m}{n}$$ $$= \displaystyle\sum_{n = 1}^{\infty} \dfrac{1}{n} \dfrac{2^{-n}}{1 - 2^{-n}}$$ $$= \displaystyle\sum_{n = 1}^{\infty} \dfrac{1}{n(2^n-1)} = A$$.

Since $$\ln B = A$$, we have $$B = e^A$$, as desired.

Because all terms are positive, every sum above is absolutely convergent, so all the steps are justified. · 4 years ago

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Can you please explain how you introduced the second summation? Thank you! · 4 years ago

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I used the Taylor series $$-\ln(1-x) = \displaystyle\sum_{n = 1}^{\infty}\dfrac{x^n}{n}$$ which is absolutely convergent for $$|x| < 1$$. · 4 years ago

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Thanks! That's a very nice solution! · 4 years ago

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Nice, thanks Jimmy! · 4 years ago

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