FOUR CUTE INFINITE SUMS FROM @infseriesbot

This is a short post, to proof 4 relations that I saw in @infseriesbot , namely

To prove these relations, recall the geometric sum (proved in the appendix of this post):

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

If we differentiate (1) with respect to multiple times we get

$f^{\prime}(x)=\sum_{k=0}^{\infty}k x^{k-1}=\frac{1}{(1-x)^2}$ (2)

$f^{\prime \prime}(x)=\sum_{k=0}^{\infty}k(k-1) x^{k-2}=\frac{2!}{(1-x)^3}$ (3)

$f^{(3)}(x)=\sum_{k=0}^{\infty}k(k-1)(k-2) x^{k-3}=\frac{3!}{(1-x)^4}$ (4)

$f^{(4)}(x)=\sum_{k=0}^{\infty}k(k-1)(k-2)(k-3) x^{k-4}=\frac{4!}{(1-x)^5}$ (5)

$\cdots$

Consider (2) and multiply it by x^2 , we get

$x^2f^{\prime}(x)=\sum_{k=0}^{\infty}k x^{k+1}=\frac{x^2}{(1-x)^2}$

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

$\sum_{k=0}^{\infty}\frac{k}{2^{k+1}}=\frac{\frac{1}{2}^2}{(1-\frac{1}{2})^2}$

$\boxed{\sum_{k=0}^{\infty}\frac{k}{2^{k+1}}=1}$

and the first result is proved.

now, multiply (2) by x^2 and (3) by x^3

$x^2f^{\prime}(x)=\sum_{k=0}^{\infty}k x^{k+1}=\frac{x^2}{(1-x)^2}$ (6)

$x^3f^{\prime \prime}(x)=\sum_{k=0}^{\infty}k(k-1) x^{k+1}=\frac{2x^3}{(1-x)^3}$ (7)

adding (6) + (7) we get:

$\sum_{k=0}^{\infty}\left(k(k-1)+k \right) x^{k+1}=\frac{x^2}{(1-x)^2}+\frac{2x^3}{(1-x)^3}$

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

$\sum_{k=0}^{\infty}\frac{k^2}{2^{k+1}}=\frac{\frac{1}{2}^2}{(1-\frac{1}{2})^2}+\frac{2\left(\frac{1}{2}\right)^3}{\left(1-\frac{1}{2}\right)^3}$

$\sum_{k=0}^{\infty}\frac{k^2}{2^{k+1}}=1+2$

$\boxed{\sum_{k=0}^{\infty}\frac{k^2}{2^{k+1}}=3}$

Now, multiply (4) by x^4

$x^4f^{(3)}(x)=\sum_{k=0}^{\infty}k(k-1)(k-2) x^{k+1}=\frac{2\cdot3 \cdot x^4}{(1-x)^4}$

$\sum_{k=0}^{\infty}\left(k^3-3k^2+2k \right) x^{k+1}=\frac{2\cdot3 \cdot x^4}{(1-x)^4}$

$\sum_{k=0}^{\infty}k^3x^{k+1}-3\sum_{k=0}^{\infty}k^2x^{k+1}+2\sum_{k=0}^{\infty}kx^{k+1} =\frac{2\cdot3 \cdot x^4}{(1-x)^4}$

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

$\sum_{k=0}^{\infty}\frac{k^3}{2^{k+1}}-3\sum_{k=0}^{\infty}\frac {k^2}{2^{k+1}}+2\sum_{k=0}^{\infty}\frac{k}{2^{k+1}} =6$

from the result above we have

$\sum_{k=0}^{\infty}\frac{k^3}{2^{k+1}}-(3\cdot 3)+ (2 \cdot 1) = 6$

$\boxed{\sum_{k=0}^{\infty}\frac{k^3}{2^{k+1}}=13}$

and the third result is proved.

For the last relation we have:

$f^{(4)}(x)=\sum_{k=0}^{\infty}k(k-1)(k-2)(k-3) x^{k-4}=\frac{4!}{(1-x)^5}$

$f^{(4)}(x)=\sum_{k=0}^{\infty}\left(k^4-6k^3+11k^2-6k \right) x^{k-4}=\frac{4!}{(1-x)^5}$

multiply the above equation by x^5

$x^5f^{(4)}(x)=\sum_{k=0}^{\infty}\ k^4x^{k+1}-6\sum_{k=0}^{\infty}k^3x^{k+1}+11\sum_{k=0}^{\infty}k^2x^{k+1}-6\sum_{k=0}^{\infty}k x^{k+1}=\frac{x^5 \cdot 4!}{(1-x)^5}$

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

$\sum_{k=0}^{\infty} \frac{k^4}{2^{k+1}}-6\sum_{k=0}^{\infty}\frac{k^3}{2^{k+1}}+11\sum_{k=0}^{\infty}\frac{k^2}{2^{k+1}}-6\sum_{k=0}^{\infty}\frac{k }{2^{k+1}}=\frac{\left( \frac{1}{2}\right)^5 \cdot 4!}{\left(1-\frac{1}{2}\right)^5}$