I have to find the sum of the series if $0

Question

I have to find this using either a power series or Taylor series method. The furthest I got was splitting it to $a_n = n^3$ and $f_n = x^n$.

Answer 1

Hint : Define $G[\lambda]=\sum_{n=0}^{\infty}e^{i\lambda n}_{}x^{n}_{}=\frac{1}{1-x e^{i\lambda}}$ and note $\sum_{n=0}^{\infty}n^{3}_{}x^{n}_{}=\frac{\partial^{3}_{}}{\partial(i\lambda)^{3}_{}}G[\lambda]\Big|_{\lambda=0}$.

Answer 2

The sum $S"'= \sum_{n=0}^{\infty}n^3 x^n$ you can think about it as the third derivative of $S= \sum_{n=0}^{\infty}x^n$ so knowing that $S=\frac{1-x^n}{1-x}$ then you do the rest.

Answer 3

One may use the following form of stars and bars: $$ \frac{1}{(1-x)^{k+1}}=\sum_{n\geq 0}\binom{n+k}{k} x^n, $$ repeated differentiation or the following approach.

$q(n)=n^3$ is a third degree polynomial in the $n$ variable. By denoting through $\delta$ the forward difference operator, $(\delta p)(n)=p(n+1)-p(n)$, we have that $\delta^4 q(n)\equiv 0$, hence $$ (1-x)^4 \sum_{n\geq 0}n^3 x^n $$ is a polynomial in the $x$ variable with degree $\leq 3$, since $$ (1-x)\sum_{n\geq 0} r(n) x^n = \sum_{n\geq 0}r(n) x^n-\sum_{n\geq 0}r(n) x^{n+1}=r(0)+\sum_{n\geq 0}(\delta r)(n)x^{n+1}. $$ By computing in a explicit way the product between $(1-x)^4$ and $0^3+1^3 x+2^3 x^2+3^3 x^3$ we get $$ (1-x)^4\sum_{n\geq 0}n^3 x^n = \left(x+4 x^2+x^3\right) $$ and we are done: $\sum_{n\geq 0}n^3 x^n=\frac{x(1+4x+x^2)}{(1-x)^4}.$

Answer 4

Hint: $n^3=n(n-1)(n-2)+3n(n-1)+n$, so $$ \begin{align} \sum_{n=0}^\infty n^3x^n &=x^3\sum_{n=0}^\infty n(n-1)(n-2)x^{n-3}+3x^2\sum_{n=0}^\infty n(n-1)x^{n-2}+x\sum_{n=0}^\infty nx^{n-1}\\ %&=\frac{6x^3}{(1-x)^4}+\frac{6x^2}{(1-x)^3}+\frac x{(1-x)^2}\\[3pt] %&=\frac{x+4x^2+x^3}{(1-x)^4} \end{align} $$