Evaluate the sum of 2k / 2^k for k from 1 to infinity.

Step-by-step solution

1. 1

   Pull the constant out and name the shape. A sum whose terms are a linear factor times a geometric factor is called arithmetico-geometric; it always converges when the ratio is under 1 in absolute value:

   $\sum_{k=1}^{\infty}\frac{2k}{2^{k}}=2\sum_{k=1}^{\infty}k\left(\frac12\right)^{k}$

2. 2

   Derive the key formula by differentiating the geometric series. From $\sum_{k\ge 0}x^{k}=\dfrac{1}{1-x}$, differentiate and multiply by $x$:

   $\sum_{k=1}^{\infty}kx^{k-1}=\frac{1}{(1-x)^{2}}\ \Longrightarrow\ \sum_{k=1}^{\infty}kx^{k}=\frac{x}{(1-x)^{2}},\qquad |x|<1$

   Differentiation is what produces the extra factor $k$; this is the standard way to generate such formulas rather than memorise them.

3. 3

   Substitute $x=\tfrac12$.

   $\sum_{k=1}^{\infty}k\left(\frac12\right)^{k}=\frac{\frac12}{\left(1-\frac12\right)^{2}}=\frac{\frac12}{\frac14}=2$

   Squaring the denominator is essential: using $1-\tfrac12$ instead of $\left(1-\tfrac12\right)^{2}$ would give 1.

4. 4

   Multiply by the constant 2.

   $\sum_{k=1}^{\infty}\frac{2k}{2^{k}}=2\times 2=4$

5. 5

   Check by adding terms. The first few are $1,1,0.75,0.5,0.3125,\dots$ and the running total climbs $1,2,2.75,3.25,3.5625,\dots$ toward $4$; summing 200 terms gives exactly $4.0$ in double precision ✓. Convergence is fast because the geometric factor eventually overwhelms the linear one.