Show that
![\[ T_n \left( \frac{x}{1-x^2} \right) = \sum_{k=0}^n x^{2k+1}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-d1d675b8cbfc60c0b9639ce8447f0698_l3.png "Rendered by QuickLaTeX.com")
Starting with the algebraic identity
![\begin{align*} &&\frac{1}{1-x^2} &= 1 + x^2 + x^4 + \cdots + x^{2n} + \frac{x^{2n+2}}{1-x^2} \\[9pt] \implies && \frac{x}{1-x^2} &= x + x^3 + x^5 + \cdots + x^{2n+1} + \frac{x^{2n+3}}{1-x^2} \\[9pt] &&&= P_{2n+1} (x) + x^{2n+2} \frac{x}{1-x^2}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-78df594c3846ac3677b28ebfaf712cd6_l3.png "Rendered by QuickLaTeX.com")
Since 
as 
we may apply Theorem 7.4 (page 277 of Apostol) to conclude
![\[ T_{2n+1} \left( \frac{x}{1-x^2} \right) = P_{2n+1} (x) = \sum_{k=0}^n x^{2k+1}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-41c8f499d3d1acbee05a12564506e6e9_l3.png "Rendered by QuickLaTeX.com")
Show that
Starting with the algebraic identity
Since as we may apply Theorem 7.4 (page 277 of Apostol) to conclude
May we also proceed by integrating the given taylor polynomial and substituting x^2 for that of -log(1-x)?