By term-by-term integration, find the sum sum from n=1

By term-by-term integration, find the sum $\displaystyle\sum_{n=1}^{\infty}n^2x^{n-1}$.

Step 1 · First differentiation

$$ \sum_{n=0}^{\infty}x^n = \frac1{1-x} \implies \sum_{n=1}^{\infty}nx^{n-1} = \frac1{(1-x)^2}. $$

Step 2 · Multiply by $x$, differentiate again

Multiplying by $x$: $\sum_{n=1}^{\infty}nx^n=\dfrac x{(1-x)^2}$. Differentiating: $$ \sum_{n=1}^{\infty}n^2x^{n-1} = \frac d{dx}\left[\frac x{(1-x)^2}\right] = \frac{(1-x)^2+2x(1-x)}{(1-x)^4} = \frac{1+x}{(1-x)^3}. $$

Step 3 · Verify

At $x=0$: both sides equal $1$. At $x=\frac12$: formula gives $\dfrac{3/2}{1/8}=12$; direct summation $1+2+2.25+2+1.5625+\cdots$ approaches $12$.

Summary Table

Additional Notes

• Repeated differentiation of the geometric series is the standard generating-function technique for producing polynomial weightings $n^k$; each differentiation raises the power of $k$ by one at the cost of an extra factor in the denominator.
• This same $\frac x{(1-x)^2}\to\frac{1+x}{(1-x)^3}$ pattern extends: differentiating once more (after another multiplication by $x$) would produce $\sum n^3x^{n-1}$, useful in combinatorial generating-function arguments.
• These generating functions appear throughout combinatorics for counting problems (e.g. the number of ways to distribute indistinguishable objects), where $n^2$ or higher weightings arise from variance-type calculations.