Using sum from n=0 to infinity x^n=frac11-x (|x|<1), find
Using $\displaystyle\sum_{n=0}^{\infty}x^n=\frac1{1-x}$ ($|x|<1$), find the sums of $1^{\circ}\ 1+2x+3x^2+4x^3+\cdots$ and $2^{\circ}\ 1+3x+6x^2+10x^3+\cdots$
Explanation of Method
Both series are obtained from the geometric series by one or two differentiations. Differentiating $\sum x^n=1/(1-x)$ once gives the first series ($n=1,2,3,\dots$ coefficients). The second series has triangular-number coefficients $\binom{n+1}2=\tfrac{n(n+1)}2$, which combine the first-derivative and second-derivative results.
- Differentiate $\sum x^n=1/(1-x)$ once to get $\sum nx^{n-1}=1/(1-x)^2$.
- Differentiate again (via $\sum nx^n$) to get $\sum n^2x^{n-1}$.
- Combine using $n(n+1)/2=(n^2+n)/2$ to find the triangular-number series.
Step 1 · Part $1^{\circ}$: differentiate once
Differentiating $\displaystyle\sum_{n=0}^{\infty}x^n=\frac1{1-x}$ term by term (valid inside the interval of convergence): $$ \sum_{n=1}^{\infty}nx^{n-1} = \frac{d}{dx}\left[\frac1{1-x}\right] = \frac1{(1-x)^2}. $$ So $$ \boxed{1+2x+3x^2+4x^3+\cdots = \frac1{(1-x)^2}}, \quad |x|<1. $$
Step 2 · Part $2^{\circ}$: build $\sum n^2x^{n-1}$
Multiply Step 1's identity by $x$: $\sum_{n=1}^{\infty}nx^n=\dfrac x{(1-x)^2}$. Differentiate again: $$ \sum_{n=1}^{\infty}n^2x^{n-1} = \frac{d}{dx}\left[\frac x{(1-x)^2}\right] = \frac{(1-x)^2+2x(1-x)}{(1-x)^4} = \frac{1+x}{(1-x)^3}. $$
Step 3 · Combine via $n(n+1)/2$
The coefficients $1,3,6,10,\dots$ are triangular numbers $T_n=\dfrac{n(n+1)}2$, so the target series is $\sum_{n=1}^{\infty}\dfrac{n(n+1)}2x^{n-1}=\dfrac12\left[\sum n^2x^{n-1}+\sum nx^{n-1}\right]$: $$ = \frac12\left[\frac{1+x}{(1-x)^3}+\frac{1-x}{(1-x)^3}\right] = \frac12\cdot\frac{2}{(1-x)^3} = \frac1{(1-x)^3}. $$ So $$ \boxed{1+3x+6x^2+10x^3+\cdots = \frac1{(1-x)^3}}, \quad |x|<1. $$
Summary Table
| Series | Closed form |
|---|---|
| $\sum nx^{n-1}$ | $1/(1-x)^2$ |
| $\sum n^2x^{n-1}$ | $(1+x)/(1-x)^3$ |
| $\sum\binom{n+1}2x^{n-1}$ | $1/(1-x)^3$ |
Additional Notes
- Both series converge on $|x|<1$ — the same radius as the parent geometric series — since term-by-term differentiation never shrinks the radius of convergence, only possibly affects behavior at the boundary $|x|=1$.
- Numerical check at $x=1/2$: Part $1^{\circ}$ gives $1/(1/2)^2=4$, matching $1+1+0.75+0.5+\cdots\to4$; Part $2^{\circ}$ gives $1/(1/2)^3=8$, matching $1+1.5+1.5+1.25+\cdots\to8$.
- This differentiation technique is the direct power-series analogue of Problem 219's term-by-term integration results: both rely on the same uniform-convergence machinery (here applied twice in succession) to justify swapping an infinite sum with a limiting operation.