Using the equality sum from n=0 to infinity frac12^n=2
Using the equality $\displaystyle\sum_{n=0}^{\infty} \frac1{2^n}=2$, find the sum of the series $\displaystyle\sum_{n=1}^{\infty} \frac n{2^{n+1}}$.
Explanation of Method
Differentiating the geometric series $\sum x^n=1/(1-x)$ term by term produces the standard identity $\sum nx^{n-1}=1/(1-x)^2$, valid for $|x|<1$. Multiplying by $x$ and evaluating at $x=\frac12$ gives $\sum n(1/2)^n$ directly in closed form, and the target series is just half of that.
- Differentiate $\sum x^n=1/(1-x)$ to get $\sum nx^{n-1}=1/(1-x)^2$.
- Multiply by $x$ and substitute $x=\frac12$.
- Relate the result to the target series via a factor of $\frac12$.
Step 1 · Differentiate the geometric series
For $|x|<1$: $\displaystyle\sum_{n=0}^{\infty}x^n=\frac1{1-x}\ \Longrightarrow\ \sum_{n=1}^{\infty}nx^{n-1}=\frac1{(1-x)^2}\ \Longrightarrow\ \sum_{n=1}^{\infty}nx^n=\frac x{(1-x)^2}.$
Step 2 · Substitute $x=\frac12$
$$ \sum_{n=1}^{\infty}n\left(\frac12\right)^n = \frac{1/2}{(1/2)^2} = 2. $$
Step 3 · Relate to the target sum
$\dfrac n{2^{n+1}}=\dfrac12\cdot n\left(\dfrac12\right)^n$, so $$ \sum_{n=1}^{\infty}\frac n{2^{n+1}} = \frac12\sum_{n=1}^{\infty}n\left(\frac12\right)^n = \frac12\cdot2 = 1. $$
Summary Table
| Quantity | Value |
|---|---|
| $\sum n(1/2)^n$ | $2$ |
| $\sum n/2^{n+1}$ | $1$ |
Additional Notes
- Term-by-term differentiation of a power series is valid within its radius of convergence (here, all of $|x|<1$), a standard fact used repeatedly throughout Problems 143–145 to bootstrap sums involving $n$, $n^2$, and $n(n-1)$ from the basic geometric series.
- Numerically, the partial sums $0.25, 0.5, 0.6875, 0.8125,\ldots$ climb steadily toward $1$, consistent with the geometric-type decay rate $\frac12$ per term.
- This problem is the direct predecessor to Problems 145's more elaborate Cauchy-product derivation of $\sum n^2/2^n=6$: the same differentiation technique, applied twice, yields all "polynomial-weighted" geometric series in this family.
Result
The series sums to: