Sequence and series (5)

Consider the infinite geometric sequence for $\lvert x\rvert <1$ :

$\frac{1}{1-x} = \sum_{k=0}^{\infty} x^k$

Differentiating both sides:

$\frac{1}{(1-x)^2} = \sum_{k=1}^{\infty} kx^{k-1}$ $\implies \frac{x}{(1-x)^2} = \sum_{k=1}^{\infty} kx^{k} \dots \ (1)$

Differentiating both sides again gives:

$\frac{x+1}{(1-x)^3} = \sum_{k=1}^{\infty} k^2x^{k-1}$ $\implies \frac{x(x+1)}{(1-x)^3} = \sum_{k=1}^{\infty} k^2x^{k}$

Differentiating both sides again gives:

$\dfrac{x^2+4x+1}{\left(x-1\right)^4} = \sum_{k=1}^{\infty} k^3x^{k-1}$ $\implies \dfrac{x^3+4x^2+x}{\left(x-1\right)^4} = \sum_{k=1}^{\infty} k^3x^{k} \dots \ (2)$

Adding (1) and (2):

$\frac{x}{(1-x)^2} + \dfrac{x^3+4x^2+x}{\left(x-1\right)^4} = \sum_{k=1}^{\infty} \left(k^3+k\right)x^{k}$

Putting $x=\frac{1}{2}$

$\frac{0.5}{(1-0.5)^2} + \dfrac{0.5^3+4(0.5)^2+0.5}{\left(0.5-1\right)^4} = \sum_{k=1}^{\infty} \frac{k^3+k}{2^k}$ $\implies \boxed{\sum_{k=1}^{\infty} \frac{k^3+k}{2^k} = 28}$

Consider the following infinite series for $|x| < 1$ :

$\begin{aligned} \sum_{k=0}^\infty x^k & = \frac 1{1-x} & \small \blue{\text{Differentiate both sides w.r.t. }x} \\ \sum_{k=1}^\infty kx^{k-1} & = \frac 1{(1-x)^2} & \small \blue{\text{Multiply both sides by }x} \\ \sum_{k=1}^\infty kx^k & = \frac x{(1-x)^2} & \small \blue{\text{Differentiate both sides w.r.t. }x} \\ \sum_{k=1}^\infty k^2x^{k-1} & = \frac {1+x}{(1-x)^3} & \small \blue{\text{Multiply both sides by }x} \\ \sum_{k=1}^\infty k^2x^k & = \frac {x+x^2}{(1-x)^3} & \small \blue{\text{Differentiate both sides w.r.t. }x} \\ \sum_{k=1}^\infty k^3x^{k-1} & = \frac {1+4x+x^2}{(1-x)^4} & \small \blue{\text{Multiply both sides by }x} \\ \sum_{k=1}^\infty k^3x^k & = \frac {x+4x^2+x^3}{(1-x)^4} \end{aligned}$

Then we have:

$\begin{aligned} \sum_{k=1}^\infty k^3x^k + \sum_{k=1}^\infty kx^k & = \frac {x+4x^2+x^3}{(1-x)^4} + \frac x{(1-x)^2} & \small \blue{\text{Putting }x=\frac 12} \\ \implies \sum_{k=1}^\infty \frac {k^3+k}{2^k} & = 26+2 = \boxed{28} \end{aligned}$

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Alternative method

$\begin{aligned} S_0 & = \sum_{k=1}^\infty \frac 1{2^k} = \frac {\frac 12}{1-\frac 12} = 1 \\ S_1 & = \sum_{\blue{k=1}}^\infty \frac k{2^k} = \sum_{\red{k=0}}^\infty \frac k{2^k} = \sum_{\red{k=0}}^\infty \frac {k+1}{2^{k+1}} = \frac 12 \blue{\sum_{\red{k=0}}^\infty \frac k{2^k}} + \sum_{\blue{k=1}}^\infty \frac 1{2^k} = \frac {\blue{S_1}}2 + S_0 \\ \implies \frac {S_1}2 & = 1 \\ S_1 & = 2 \end{aligned}$

Similarly

$\begin{aligned} S_2 & = \sum_{\blue{k=1}}^\infty \frac {k^2}{2^k} = \sum_{\red{k=0}}^\infty \frac {k^2}{2^k} = \sum_{\red{k=0}}^\infty \frac {(k+1)^2}{2^{k+1}} = \frac 12 \sum_{\red{k=0}}^\infty \frac {k^2+2k + 1}{2^k} \\ & = \frac {S_2}2 + S_1 + S_0 = \frac {S_2}2 + 2 + 1 \\ & = 2(2+1) = 6 \end{aligned}$

And

$\begin{aligned} S_3 & = \sum_{\blue{k=1}}^\infty \frac {k^3}{2^k} = \sum_{\red{k=0}}^\infty \frac {k^3}{2^k} = \sum_{\red{k=0}}^\infty \frac {(k+1)^3}{2^{k+1}} = \frac 12 \sum_{\red{k=0}}^\infty \frac {k^3+3k^2+3k + 1}{2^k} \\ & = \frac {S_3}2 + \frac {3S_2}2 + \frac {3S_1}2 + S_0 \\ & = 2\left(9+3+1\right) = 26 \end{aligned}$

Therefore $\displaystyle \sum_{k=1}^\infty \frac {k^3+k}{2^k} = S_3 + S_1 = 26 + 2 = \boxed{28}$ .