Evaluate the sum \(\sum_{k = 1}^{\infty} (2k + 1)x^{2k}\).
We have to assume \(\vert x \vert < 1\), otherwise we can’t use the infinite series approximation.
We can calculate \(\sum_{k = 1}^{\infty} x^{2k + 1}\) as follows:
\[\begin {align} \sum_{k = 1}^{\infty} x^{2k + 1} & = x^3 + x^5 + x^7 + \cdots \\ & = x + x^3 + x^5 + x^7 + \cdots - x \\ & = x(1 + x^2 + x^4 + x^6 + \cdots) - x \\ & = x \cdot \sum_{k = 0}^{\infty} (x^2)^k - x \\ & = x \cdot \frac 1 {1 - x^2} - x \\ & = \frac {x - x(1 - x^2)} {(1 - x^2)} \\ & = \frac {x^3} {(1 - x^2)} \end {align}\]Differentiating both sides of the above equation w.r.t \(x\):
\[\begin {align} \sum_{k = 1}^{\infty} (2k + 1)x^{2k} & = \frac d {dx} \frac {x^3} {(1 - x^2)} \\ & = \frac {3x^2(1 - x^2) - x^3(-2x)} {(1 - x^2)^2} \\ & = \frac {3x^2 - 3x^4 + 2x^4} {(1 - x^2)^2} \\ & = \frac {x^2(3 - x^2)} {(1 - x^2)^2} \end {align}\]