+26393 sequence proof
\(\text{Let $\mathbf{y=1+x}$ or $\mathbf{x = y-1}$ }\)
\(\begin{array}{|rcll|} \hline f(x) &=& & \dfrac{1}{y} + \dfrac{2}{y^2} + \dfrac{3}{y^3} + \dfrac{4}{y^4} +\ldots + \dfrac{n-1}{y^{n-1}}+ \dfrac{n}{y^n} \quad | \quad \cdot y \\ yf(x) &=&1 +& \dfrac{2}{y } + \dfrac{3}{y^2} + \dfrac{4}{y^3} + \dfrac{5}{y^4} +\ldots + \dfrac{n}{y^{n-1}} \\ \hline yf(x)-f(x) &=& 1 + & \dfrac{1}{y } + \dfrac{1}{y^2} + \dfrac{1}{y^3} + \dfrac{1}{y^4} +\ldots + \dfrac{1}{y^{n-1}} - \dfrac{n}{y^n} \\ (y-1)f(x) &=& 1 + & \dfrac{1}{y } + \dfrac{1}{y^2} + \dfrac{1}{y^3} + \dfrac{1}{y^4} +\ldots + \dfrac{1}{y^{n-1}} - \dfrac{n}{y^n} \quad | \quad y-1 = x \\ xf(x) &=& 1 + & \dfrac{1}{y } + \dfrac{1}{y^2} + \dfrac{1}{y^3} + \dfrac{1}{y^4} +\ldots + \dfrac{1}{y^{n-1}} - \dfrac{n}{y^n} \\ \hline \end{array} \)
\(\begin{array}{|rclcl|} \hline xf(x) &=& \underbrace{1 + \dfrac{1}{y } + \dfrac{1}{y^2} + \dfrac{1}{y^3} + \dfrac{1}{y^4} +\ldots + \dfrac{1}{y^{n-1}}}_{=s~(\text{sum of a geometric progression})} - \dfrac{n}{y^n} \\ xf(x) &=& \mathbf{s} - \dfrac{n}{y^n} \\ \hline && \mathbf{s} = 1 + \dfrac{1}{y } + \dfrac{1}{y^2} + \dfrac{1}{y^3} + \dfrac{1}{y^4} +\ldots + \dfrac{1}{y^{n-1}} \quad | \quad : y \\ && \dfrac{\mathbf{s}}{y} = \quad \dfrac{1}{y } + \dfrac{1}{y^2} + \dfrac{1}{y^3} + \dfrac{1}{y^4} +\ldots + \dfrac{1}{y^{n-1}} + \dfrac{1}{y^{n}} \\ \hline && \mathbf{s}-\dfrac{\mathbf{s}}{y} = 1- \dfrac{1}{y^{n}} \\ && \mathbf{s}\left(1-\dfrac{1}{y}\right) = 1- \dfrac{1}{y^{n}} \\ && \mathbf{s} = \dfrac{1- \dfrac{1}{y^{n}}}{1-\dfrac{1}{y}} \\ \hline xf(x) &=& \mathbf{\dfrac{1- \dfrac{1}{y^{n}}}{1-\dfrac{1}{y}}} - \dfrac{n}{y^n} \quad | \quad y=1+x \\ \mathbf{xf(x)} &=& \mathbf{\dfrac{1- \dfrac{1}{(1+x)^{n}}}{1-\dfrac{1}{1+x}} - \dfrac{n}{(1+x)^n} } \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{xf(x)} &=& \mathbf{\dfrac{1- \dfrac{1}{(1+x)^{n}}}{1-\dfrac{1}{1+x}} - \dfrac{n}{(1+x)^n} } \\ \\ xf(x) &=& \dfrac{ \dfrac{(1+x)^{n}-1}{(1+x)^{n}}} { \dfrac{1+x-1 }{1+x}} - \dfrac{n}{(1+x)^n} \\ \\ xf(x) &=& \dfrac{ \dfrac{(1+x)^{n}-1}{(1+x)^{n}}} { \dfrac{x}{1+x}} - \dfrac{n}{(1+x)^n} \\ \\ xf(x) &=& \dfrac{\Big( (1+x)^{n}-1 \Big)(1+x) } {x(1+x)^{n}} - \dfrac{n}{(1+x)^n}\cdot \dfrac{x}{x} \\ \\ xf(x) &=& \dfrac{\Big( (1+x)^{n}-1 \Big)(1+x)-nx } {x(1+x)^{n}} \quad | \quad : x \\ \\ f(x) &=& \dfrac{\Big( (1+x)^{n}-1 \Big)(1+x)-nx } {x^2(1+x)^{n}} \\ \\ \mathbf{f(x)} &=& \mathbf{ \dfrac{ (1+x)^{n+1}-(1+x)-nx } {x^2(1+x)^{n}} } \\ \hline \end{array}\)
