Compute
\(1 \cdot \dfrac {1}{2} + 2 \cdot \dfrac {1}{4} + 3 \cdot \dfrac {1}{8} + \dots + n \cdot \dfrac {1}{2^n} + \dotsb\)
\(\text{Let $x = \dfrac{1}{2} \quad|x|<1$ } \)
Now we have: \(\displaystyle s= \sum \limits_{n=1}^{\infty} nx^n \)
Let the progression to be summed be put equal to s:
\(s = x+2x^2+3x^3+4x^4+\cdots + nx^n +\cdots\)
It is divided by \(x\) and multiplied by \(dx\) then
\(\dfrac{s\ dx}{x} = dx +2x\ dx+3x^2\ dx+4x^3\ dx+\cdots + nx^{n-1}\ dx +\cdots \)
and with the integrals taken this equation is found
\(\displaystyle \int \dfrac{s\ dx}{x} = x +x^2+x^3+x^4 +\cdots + x^n + \cdots = \dfrac{x}{1-x} \quad \text{ infinite geometric progression }\)
Therefore from the equation:
\(\displaystyle \int \dfrac{s\ dx}{x} = \dfrac{x}{1-x}\)
on differentiation s is found. For the equation becomes:
\(\begin{array}{rcll} \displaystyle \dfrac{s}{x} &=& \dfrac{x}{1-x}\left( \dfrac{1}{x}- \dfrac{(-1)}{1-x} \right) \\\\ &=& \dfrac{1}{1-x} + x(-1)(1-x)^{-2}(-1) \\\\ &=& \dfrac{1}{1-x} + \dfrac{x}{(1-x)^2} \\\\ &=& \dfrac{1-x+x}{(1-x)^2} \\\\ &=& \dfrac{1 }{(1-x)^2} \end{array}\)
thus there is produced:
\(\displaystyle s = \dfrac{x}{(1-x)^2} \\\)
So \(\text{let $x = \dfrac{1}{2} $}:\)
\(\begin{array}{|rcll|} \hline s &=& \dfrac{ \dfrac{1}{2} } {\left(1-\dfrac{1}{2} \right)^2} \\\\ &=& \dfrac{ \dfrac{1}{2} } {\left( \dfrac{1}{2} \right)^2} \\\\ &=& \dfrac{ 1 } { \dfrac{1}{2} } \\\\ \mathbf{s} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)
finally
\(\displaystyle 1 \cdot \dfrac {1}{2} + 2 \cdot \dfrac {1}{4} + 3 \cdot \dfrac {1}{8} + \dots + n \cdot \dfrac {1}{2^n} + \dotsb = 2\)
