Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? [duplicate]

Solution 1:

Since infinite series with nonnegative terms can be rearranged arbitrarily,

$$\sum_{i=1}^\infty \frac{i}{2^i} = \sum_{i=1}^\infty \sum_{j=1}^i \frac{1}{2^i} = \sum_{j=1}^\infty \sum_{i=j}^\infty \frac{1}{2^i} = \sum_{j=1}^\infty \frac{1}{2^{j-1}} = 2 $$

More graphically,

  1/2 + 2/4 + 3/8 + 4/16 + ...

= 1/2 + 1/4 + 1/8 + 1/16 + ...    (= 1)
      + 1/4 + 1/8 + 1/16 + ...    (= 1/2)           
            + 1/8 + 1/16 + ...    (= 1/4)
                  + 1/16 + ...    (= 1/8)
                        ....       ....

Solution 2:

\begin{gather*} |x|<1:\quad f(x)=\sum_{n=1}^{\infty} x^n=\frac{x}{1-x} \\ xf'(x)=\sum_{n=1}^{\infty} nx^n=\frac{x}{(1-x)^2} \end{gather*}

Let $x=\frac{1}{2}$

Solution 3:

Let $s = 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + \dots$

Then $2s = 1 + 2/2 + 3/4 + 4/8 + 5/16 + \dots$

And then subtracting terms with similar denominators gives: $2s - s = 1 +1/2+1/4+1/8+1/16+\dots = 2$