\(\[1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.\]\)Compute
it's so urgent you couldn't post the image so we could see it!
\(\[1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.\]\)
∑[n/(2^n), n, 1, ∞] =2
Here is the partial sum formula:
sum_(n=1 to m) = n/2^n = 2^(-m) (-m + 2^(m + 1) - 2)
Here, we can use \(\frac{n}{1-r}\) , so \(\frac{1}{1-\frac{1}{2}}=\frac{1}{\frac{1}{2}}=\boxed{2}\) .
