Find the set of convergence and the sum of the following power series: $\sum_{n=1}^{\infty} n^3x^n $

Question

This is the exercise: $$\sum_{n=1}^{\infty} n^3x^n.$$

I only managed to find the the set of convergence to be $(-1,1)$. The answer should be: $$\sum_{n=1}^{\infty} n^3x^n =\frac{x(x^2+4x+1)}{(1-x)^4}.$$

Answer 1

From sum of geometric series we know that $\sum_{n=0}^\infty x^n=\frac{1}{1-x}$ when $x\in(-1,1)$. From a known theorem about power series we can take the derivative on both sides and get $\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$. Multiply both sides by $x$ to get $\sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2}$. Now you can take a derivative on both sides again, and continue this way. It is a bit of work but it will get you to the result you need.

Answer 2

Hint: Given that $$(1-x)^{-1}=\sum_{n\geq 0}x^n,$$ You differentiate the equation once to get $$-(1-x)^{-2}=\sum_{n\geq 0}nx^{n-1},$$ And multiplying both sides by $x$ you get $$-x(1-x)^{-2}=\sum_{n\geq 0}nx^{n}$$ Now you differentiate again, and then again...

Answer 3

some ingredients: $$ (1-x)^{-2} = \left( \begin{array}{c}1\\1 \end{array}\right) + \left( \begin{array}{c}2\\1 \end{array}\right) x +\left( \begin{array}{c}3\\1 \end{array}\right) x^2 +\left( \begin{array}{c}4\\1 \end{array}\right) x^3 +\left( \begin{array}{c}5\\1 \end{array}\right) x^4 +\left( \begin{array}{c}6\\1 \end{array}\right) x^5 + \cdots $$

$$ (1-x)^{-3} = \left( \begin{array}{c}2\\2 \end{array}\right) + \left( \begin{array}{c}3\\2 \end{array}\right) x +\left( \begin{array}{c}4\\2 \end{array}\right) x^2 +\left( \begin{array}{c}5\\2 \end{array}\right) x^3 +\left( \begin{array}{c}6\\2 \end{array}\right) x^4 +\left( \begin{array}{c}7\\2 \end{array}\right) x^5 + \cdots $$

$$ (1-x)^{-4} = \left( \begin{array}{c}3\\3 \end{array}\right) + \left( \begin{array}{c}4\\3 \end{array}\right) x +\left( \begin{array}{c}5\\3 \end{array}\right) x^2 +\left( \begin{array}{c}6\\3 \end{array}\right) x^3 +\left( \begin{array}{c}7\\3 \end{array}\right) x^4 +\left( \begin{array}{c}8\\3 \end{array}\right) x^5 + \cdots $$

Answer 4

Similar basis, but a different strategy: as $\;\sum_{n\ge 0}x^n=\dfrac1{1-x}$ the successive derivatives yield:

\begin{alignat}{2} &\bullet\enspace\Bigl(\sum_{n\ge 0}x^n\Bigr)'=\sum_{n\ge 1}n\, x^{n-1}=\frac1{(1-x)^2},&\text{so}\quad& \sum_{n\ge 1}n\, x^{n}=\frac x{(1-x)^2},\\ &\bullet\enspace\Bigl(\sum_{n\ge 1}x^n\Bigr)''=\sum_{n\ge 1}n(n-1)\, x^{n-2}=\frac2{(1-x)^3},&\quad\text{so}\quad&\sum_{n\ge 1}n(n-1)\, x^{n}=\frac{2x^2}{(1-x)^3}, \\ &\bullet\enspace\Bigl(\sum_{n\ge 1}x^n\Bigr)'''=\sum_{n\ge 1}n(n-1)(n-2)\, x^{n-2}=\frac6{(1-x)^4},&\quad\text{so}\quad&\sum_{n\ge 1}n(n-1)(n-2)\, x^{n}=\frac{6x^3}{(1-x)^4}. \end{alignat} Now $n^3=n(n-1)(n-2)+3n^2-2n$, $\quad n^2=n(n-1)+n$, therefore $$n^3=n(n-1)(n-2)+3n(n-1)+n, $$ so that by linearity, $$\sum_{n\ge 1}n^3x^n=\frac{6x^3}{(1-x)^4}+\frac{6x^2}{(1-x)^3}+\frac{x}{(1-x)^2}.$$