# Calculation of an infinite sum $\sum_{k=1}^{\infty} \frac{k^3}{2^k} = 26$ [closed]

Could you give me some advice on how to calculate the following sum? \begin{aligned}\sum_{k=1}^{\infty} \frac{k^3}{2^k} = 26\end{aligned}

Thank you!

## closed as off-topic by Stella Biderman, apnorton, Henrik, Claude Leibovici, user223391 Oct 10 '16 at 22:32

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$$S(x)=\sum_{k=1}^\infty k^3x^k\tag{x=\frac12}$$

$$\int_0^x\frac1{t_1}\int_0^{t_1}\frac1{t_2}\int_0^{t_2}\frac{S(t_3)}{t_3}\ dt_3\ dt_2\ dt_1 =\int_0^x\frac1{t_1}\int_0^{t_1}\frac1{t_2}\int_0^{t_2}\sum_{k=1}^\infty k^3(t_3)^{k-1}\ dt_3\ dt_2\ dt_1$$ \begin{align} & =\sum_{k=1}^\infty\int_0^x\frac1{t_1}\int_0^{t_1}\frac1{t_2}\int_0^{t_2}k^3(t_3)^{k-1}\ dt_3\ dt_2\ dt_1 \\ & =\sum_{k=1}^\infty\int_0^x\frac1{t_1}\int_0^{t_1}k^2(t_2)^{k-1}\ dt_2\ dt_1 \\ & =\sum_{k=1}^\infty\int_0^xk(t_1)^{k-1}\ dt_1 \\ & =\sum_{k=1}^\infty x^k \\ & =\frac x{1-x}\tag{geometric series} \end{align}

$$\int_0^x\frac1{t_1}\int_0^{t_1}\frac1{t_2}\int_0^{t_2}\frac{S(t_3)}{t_3}\ dt_3\ dt_2\ dt_1=\frac x{1-x}$$

$$S(x)=x\left(\frac{d}{dx}x\left(\frac{d}{dx}x\left(\frac{d}{dx}\frac x{1-x}\right)\right)\right)$$

Then consider $x=\frac12$.

\begin{aligned} & S_{0} = \sum_{n=1}^{\infty} \frac{n^{0}}{2^{n}} = \sum_{n=1}^{\infty} \frac{1}{2^{n}} \Rightarrow \color{red}{S_{0} = 1} \\ \\ & S_{1} = \sum_{n=1}^{\infty} \frac{n^{1}}{2^{n}} = \sum_{n=1}^{\infty} \frac{n + 1 - 1}{2^{n}} = \sum_{n=1}^{\infty} \frac{n + 1}{2^{n}} - S_{0} \\ & \qquad \Rightarrow S_{1} + S_{0} = \sum_{n=1}^{\infty} \frac{n + 1}{2^{n}} = 2 \sum_{n=1}^{\infty} \frac{n + 1}{2^{n + 1}} = 2 \sum_{n=2}^{\infty} \frac{n}{2^{n}} = 2 \left[ - \frac{1}{2} + \sum_{n=1}^{\infty} \frac{n}{2^{n}} \right] = - 1 + 2 S_{1} \\ & \qquad \Rightarrow \color{red}{S_{1} = 1 (S_{0} + 1) = 2} \\ \\ & S_{2} = \sum_{n=1}^{\infty} \frac{n^{2}}{2^{n}} = \sum_{n=1}^{\infty} \frac{n^{2} + 2 n + 1 - 2 n - 1}{2^{n}} = \sum_{n=1}^{\infty} \frac{(n + 1)^{2}}{2^{n}} - 2 S_{1} - S_{0} \\ & \qquad \Rightarrow S_{2} + 2 S_{1} + S_{0} = S_{2} + 3 (S_{0} + 1) - 1 = 2 \sum_{n=1}^{\infty} \frac{(n + 1)^{2}}{2^{n + 1}} = - 1 + 2 S_{2} \\ & \qquad \Rightarrow \color{red}{S_{2} = 3 (S_{0} + 1) = 6} \\ \\ & S_{3} = \sum_{n=1}^{\infty} \frac{n^{3}}{2^{n}} = \sum_{n=1}^{\infty} \frac{(n + 1)^{3} - 3 n^{2} - 3 n - 1}{2^{n}} = \sum_{n=1}^{\infty} \frac{(n + 1)^{3}}{2^{n}} - 3 S_{2} - 3 S_{1} - S_{0} \\ & \qquad \Rightarrow S_{3} + 3 S_{2} + 3 S_{1} + S_{0} = S_{3} + 13 (S_{0} + 1) - 1 = 2 \sum_{n=1}^{\infty} \frac{(n + 1)^{2}}{2^{n + 1}} = - 1 + 2 S_{3} \\ & \qquad \Rightarrow \color{red}{S_{3} = 13 (S_{0} + 1) = 26} \\ \\ & \text{... etc} \end{aligned}

I will add what commented previously. We have that

$$\Delta x^{\underline{k}}=kx^{\underline {k-1}}\quad\text{and}\quad\sum a^{k}\delta k=\frac{a^k}{a-1}+C\tag{1}$$

where $x^{\underline k}$ is a falling factorial. And the formula for summation by parts

$$\sum f\Delta g=fg-\sum {\rm E}g\Delta f\tag{2}$$

where $\Delta f(k)=f(k+1)-f(k)$ and ${\rm E}f(k)=f(k+1)$. And we can write any monomial as

$$x^n=\sum_{k=1}^n \left\{{n\atop k}\right\}x^{\underline k}$$

where $\left\{{n\atop 1}\right\}=\left\{{n\atop n}\right\}=1$ and $\left\{{n\atop 2}\right\}=2^{n-1}-1$ for any $n\in\Bbb N$. Then

$$k^3=k+3k^{\underline 2}+k^{\underline 3}$$

Now setting $f(k)=k^3$ and $\Delta g(k)=\left(\frac12\right)^k$ and using $(2)$ and $(1)$ we get

$$\sum\frac{k^3}{2^k}\delta k=-\frac{k^3}{2^{k-1}}+\sum \frac{1+6k+3k^{\underline 2}}{2^k}\delta k$$

Now applying summation by parts again with $f(k)=1+6k+3k^{\underline 2}$ and $\Delta g(k)=\frac1{2^k}$ we get

$$\sum\frac{k^3}{2^k}\delta k=-\frac{k^3}{2^{k-1}}+ \left(-\frac{1+6k+3k^{\underline 2}}{2^{k-1}}+\sum\frac{6+6k}{2^k}\delta k\right)$$

and repeating again summation by parts we get finally

$$\sum\frac{k^3}{2^k}\delta k=-\frac{k^3}{2^{k-1}}-\frac{1+6k+3k^{\underline 2}}{2^{k-1}}-\frac{6+6k}{2^{k-1}}-\frac6{2^{k-1}}+C$$

Then taking limits

$$\sum_{k\ge 1}\frac{k^3}{2^k}=\sum\nolimits_1^{\infty}\frac{k^3}{2^k}\delta k=-\frac1{2^{k-1}}(k^3+9k+13+3k^2)\Big|_1^{\infty}=26$$

Maybe a more faster calculation is using the recursive formula for summation by parts

$$\sum f \Delta g=\sum_{k\ge 0}(-1)^k\ \Delta ^k f\ \frac{{\rm E}^k}{\Delta^k}g$$