# Calculate sum_{n=0}^{infty}n^3((x+1)/2)^n ?

Jun 13, 2016

$f \left(x\right) = \frac{2 \left(1 + x\right) \left(13 + 10 x + {x}^{2}\right)}{x - 1} ^ 4$ for $- 3 < x < 1$

#### Explanation:

The target series is $S = {\sum}_{n = 0}^{\infty} {n}^{3} {\left(\frac{x + 1}{2}\right)}^{n}$ with generic term
$t = {n}^{3} {\left(\frac{x + 1}{2}\right)}^{n}$

Let us introduce ${S}_{b} = {\sum}_{n = 0}^{\infty} {\left(\frac{x + 1}{2}\right)}^{n}$ with generic term ${t}_{b} = {\left(\frac{x + 1}{2}\right)}^{n}$

${S}_{b}$ is convergent for $- 3 < x < 1$. Comparing terms we have

${D}_{3} = {d}^{3} / \left({\mathrm{dx}}^{3}\right) {t}_{b} = \frac{n \left(n - 1\right) \left(n - 2\right)}{{2}^{3}} {\left(\frac{x + 1}{2}\right)}^{n - 3} =$
${n}^{3} / \left({2}^{3}\right) {\left(\frac{x + 1}{2}\right)}^{n - 3} - \frac{3 {n}^{2}}{{2}^{3}} {\left(\frac{x + 1}{2}\right)}^{n - 3} + \frac{2 n}{{2}^{3}} {\left(\frac{x + 1}{2}\right)}^{n - 3}$

then

${n}^{3} {\left(\frac{x + 1}{2}\right)}^{n} = {2}^{3} {\left(\frac{x + 1}{2}\right)}^{3} {d}^{3} / \left({\mathrm{dx}}^{3}\right) t + 3 {n}^{2} {\left(\frac{x + 1}{2}\right)}^{n} - 2 n {\left(\frac{x + 1}{2}\right)}^{n}$

Also

${D}_{2} = {d}^{2} / \left({\mathrm{dx}}^{2}\right) {t}_{b} = \frac{n \left(n - 1\right)}{n} {\left(\frac{x + 1}{2}\right)}^{n - 2}$ and analogously

${2}^{2} {\left(\frac{x + 1}{2}\right)}^{2} {d}^{2} / \left({\mathrm{dx}}^{2}\right) {t}_{b} = {n}^{2} {\left(\frac{x + 1}{2}\right)}^{n} - n {\left(\frac{x + 1}{2}\right)}^{n}$

$2 \left(\frac{x + 1}{2}\right) \frac{d}{\mathrm{dx}} {t}_{b} = n {\left(\frac{x + 1}{2}\right)}^{n}$

$2 \left(\frac{x + 1}{2}\right) {D}_{1} = n {\left(\frac{x + 1}{2}\right)}^{n}$

Putting all together

${S}_{0} = {\sum}_{n = 0}^{\infty} \left\{{2}^{3} {\left(\frac{x + 1}{2}\right)}^{3} {D}_{3} + 3 \times {2}^{2} {\left(\frac{x + 1}{2}\right)}^{2} {D}_{2} + 2 \left(\frac{x + 1}{2}\right) {D}_{1}\right\}$

or

${S}_{o} = \frac{2 \left(1 + x\right) \left(13 + 10 x + {x}^{2}\right)}{x - 1} ^ 4$ for $- \infty < x < 1$

Attached is a figure with the comparison between $S$ and

${S}_{0} = \frac{2 \left(1 + x\right) \left(13 + 10 x + {x}^{2}\right)}{x - 1} ^ 4$

Note:

The final expression for ${S}_{0}$ is obtained considering that ${D}_{k} , \left\{k = 1 , 2 , 3\right\}$ is applied on the equivalence S_b equiv 1/(1-((x+1)/2) for $- 3 < x < 1$

so

${D}_{1} = \frac{2}{x - 1} ^ 2$
${D}_{2} = - \frac{4}{x - 1} ^ 3$
${D}_{3} = \frac{12}{x - 1} ^ 4$