Using the Cauchy product, show that sum from n=2
Using the Cauchy product, show that $\displaystyle\sum_{n=2}^{\infty} \frac{n(n-1)}{2^{n-2}}=16$, then find the sum of the series $\displaystyle\sum_{n=1}^{\infty} \frac{n^2}{2^n}$.
Explanation of Method
Taking the Cauchy product of $\sum1/2^n$ with itself (both absolutely convergent geometric series) produces $\sum(n+1)/2^n=4$ by direct multiplication of the sums. Repeating the Cauchy product once more against the new series yields $\sum(n+1)(n+2)/2^{n+1}=8$, which reindexes exactly to the target identity. Finally, writing $n^2=n(n-1)+n$ splits $\sum n^2/2^n$ into this known result plus the simpler $\sum n/2^n$ from Problem 143.
- Take the Cauchy product of $\sum1/2^n$ with itself to get $\sum(n+1)/2^n=4$.
- Take the Cauchy product again to get $\sum(n+1)(n+2)/2^{n+1}=8$, and reindex to the target identity.
- Combine via $n^2=n(n-1)+n$ to find $\sum n^2/2^n$.
Step 1 · First Cauchy product
With $a_n=b_n=1/2^n$: $c_n=\sum_{k=0}^n\frac1{2^k}\cdot\frac1{2^{n-k}}=\dfrac{n+1}{2^n}$. Since both factors converge absolutely to $2$: $\sum_{n=0}^{\infty}\dfrac{n+1}{2^n}=2\cdot2=4$. Splitting off the constant part: $\sum_{n=1}^{\infty}n/2^n=4-2=2$ (matching Problem 143's result via $\sum n/2^{n+1}=1$).
Step 2 · Second Cauchy product and reindexing
Taking the Cauchy product of $\sum1/2^n$ (sum $2$) with $\sum(n+1)/2^n$ (sum $4$): $c_n=\dfrac1{2^n}\sum_{k=0}^n(n-k+1)=\dfrac{(n+1)(n+2)}{2^{n+1}}$, so $\sum_{n=0}^{\infty}\dfrac{(n+1)(n+2)}{2^{n+1}}=2\cdot4=8$. Multiplying by $4$ and reindexing $m=n+2$: $\sum_{m=2}^{\infty}\dfrac{m(m-1)}{2^{m-2}}=16$.
Step 3 · Combine to find $\sum n^2/2^n$
Since $n^2=n(n-1)+n$: $$ \sum_{n=1}^{\infty}\frac{n^2}{2^n}=\sum_{n=2}^{\infty}\frac{n(n-1)}{2^n}+\sum_{n=1}^{\infty}\frac n{2^n} = \frac14\cdot16+2 = 4+2=6. $$
Summary Table
| Quantity | Value |
|---|---|
| $\sum(n+1)/2^n$ | $4$ |
| $\sum n(n-1)/2^{n-2}$ | $16$ |
| $\sum n^2/2^n$ | $6$ |
Additional Notes
- This is a direct sequel to Problem 143's differentiation-based approach: both problems arrive at $\sum n/2^n=2$, but this one demonstrates an entirely different technique (Cauchy products of absolutely convergent series) reaching the same and further results, illustrating the flexibility of series manipulation tools.
- Repeated Cauchy products of $\sum x^n$ with itself generate the successive derivatives of $1/(1-x)$, so this construction is really an algebraic (rather than calculus-based) route to the same power-series identities used in Problem 143 — a useful alternative when differentiation isn't the preferred tool.
- The identity $n^2=n(n-1)+n$ is the standard decomposition trick for evaluating $\sum n^2x^n$-type series once $\sum n(n-1)x^n$ (the "second derivative" sum) and $\sum nx^n$ (the "first derivative" sum) are both known — a pattern extendable to $\sum n^3x^n$ via $n^3=n(n-1)(n-2)+3n(n-1)+n$.
Result
Both identities hold: