Puzzles
Sum
What is
$$\sum_{i=1}^{\infty}\frac{1}{i 2^i}$$
?
Show answer & extension
Hide answer & extension
Write \(x\) instead of 2. Now we find that:
$$\frac{d}{dx}\sum_{i=1}^{\infty}\frac{1}{i x^i}=\sum_{i=1}^{\infty}\frac{-1}{x^{i+1}}\\
=-\frac{1}{x}\sum_{i=1}^{\infty}\frac{1}{x^i}\\
=-\frac{1}{x}\times\frac{1}{x-1}\\
=\frac{1}{x}+\frac{1}{1-x}$$
The sum we are after is the integral of this:
$$\sum_{i=1}^{\infty}\frac{1}{i x^i}=\ln|x|+\ln|1-x|$$
When \(x=2\), this is \(\ln(2)\)
Extension
What is
$$\sum_{i=1}^{\infty}\frac{1}{i^2 2^i}$$
?
