Compute $1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \cdots + n \cdot \frac {1}{2^n} + \cdots $ I have tried to compute the first few terms to try to find a pattern but I got
$$\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}$$
but I still don't see any obvious pattern(s). I also tried to look for a pattern in the question, but I cannot see any pattern (possibly because I'm overthinking it?) Please help me with this problem.
 A: $$\sum_{n=1}^\infty nx^n=x\sum_{n=1}^\infty nx^{n-1}=x\sum_{n=0}^\infty (x^n)'=x\Bigl(\sum_{n=0}^\infty x^n\Bigr)'=x\Bigl(\frac1{1-x}\Bigr)'=\frac x{(1-x)^2}.$$
A: $$I=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}+\cdots$$
$$2I=1+1+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\frac{6}{32}+\cdots$$
$$2I-I=1+\left(1-\frac 12 \right)+\left(\frac 34 -\frac 24 \right)+\left(\frac 48 -\frac 38 \right)+\left(\frac {5}{16} -\frac {4}{16} \right)+\cdots$$
$$I=1+\frac 12+\frac 14+\frac 18+\cdots=2$$
A: Summation by parts gives:
$$ \sum_{n=1}^{N}\frac{n}{2^n} = N\left(1-\frac{1}{2^N}\right)-\sum_{n=1}^{N-1}\left(1-\frac{1}{2^k}\right)=1-\frac{N}{2^N}+\left(1-\frac{1}{2^{N-1}}\right)=\color{red}{2-\frac{N+2}{2^N}}. $$
A: Let
$$ L = \sum_{n=1}^\infty \frac{n}{2^n} $$
Then,
$$ L = \frac{1}{2} + \sum_{n=2}^\infty \frac{n}{2^n} \\  
L = \frac{1}{2} + \sum_{n=1}^\infty \frac{n+1}{2^{n+1}} \\  
L = \frac{1}{2} + \frac{1}{2} \sum_{n=1}^\infty \frac{n}{2^n} + \frac{1}{2} \sum_{n=1}^\infty \frac{1}{2^n} \\  
L = \frac{1}{2} + \frac{L}{2}  + \frac{1}{2} \sum_{n=1}^\infty \frac{1}{2^n} \\  
\frac{L}{2} = \frac{1}{2}  + \frac{1}{2} \sum_{n=1}^\infty \frac{1}{2^n} \\  
\frac{L}{2} = \frac{1}{2} + \frac{1}{2} \\  
\boxed{L = 2}
$$
A: First consider the partial sum by showing
$$ \sum_{n=1}^k \frac{n}{2^n} = 2^{-k}(-k+2^{k+1}-2). $$
Now $k\to \infty$ gives the result $2$.

Edit: Proof of the partial sum by induction
$k=1$: 
$$\sum_{n=1}^1 \frac{1}{2} = 1/2 = 2^{-1}(-1+2^{2}-2) \quad \checkmark$$
Let $\sum_{n=1}^k \frac{n}{2^n} = 2^{-k}(-k+2^{k+1}-2)$ be true for any $k\geq1$. 
Induction step:
$$ \sum_{n=1}^{k+1} \frac{n}{2^n} = \sum_{n=1}^k \frac{n}{2^n} + \frac{k+1}{2^{k+1}} = 2^{-k}(-k+2^{k+1}-2) + \frac{k+1}{2^{k+1}} = \frac{2(-k+2^{k+1}-2)+k+1}{2^{k+1}} = \frac{-k+2^{k+2}-3}{2^{k+1}} = 2^{-(k+1)}(-(k+1)+2^{k+2}-2) $$
