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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, 5 months ago

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5 votes

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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.

Jimmy Kariznov - 4 years, 5 months ago

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Can you please explain how you introduced the second summation? Thank you!

Pranav Arora - 4 years, 5 months 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 \(|x| < 1\).

Jimmy Kariznov - 4 years, 5 months ago

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@Jimmy Kariznov Thanks! That's a very nice solution!

Pranav Arora - 4 years, 5 months ago

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Nice, thanks Jimmy!

Fariz Azmi Pratama - 4 years, 5 months ago

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