Summation and Limit?

I got this nice question from my friend.

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

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

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

#Algebra #MathProblem #Math

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 - 7 years, 10 months ago

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

Pranav Arora - 7 years, 10 months ago

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 - 7 years, 10 months ago

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

Pranav Arora - 7 years, 10 months ago

Nice, thanks Jimmy!

Fariz Azmi Pratama - 7 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$