Summation by Parts to Evaluate $\sum_{k=1}^{\infty}(2k+1)x^{2k}$ I need to evaluate $\sum_{k=1}^{\infty}(2k+1)x^{2k}$ using the Summation by Parts (SBP) method. It is given that $0 < |x| < 1$.
The notation our class uses for SBP is as follows:
$$ \sum_{i} u_i\cdot\triangle v_i = u_i\cdot v_i - \sum_{i} \triangle u_i \cdot v_{i+1}$$
where $u_i$ and $v_i$ are both sequences.
I have already evaluated the sum using differentiation, but when I attempt SBP I do not get the same result.
Here are my steps for differentiation, with the correct result:
$$ \frac{d}{dx} \left(x^{2k+1}\right) = (2k+1)x^{2k} $$
                $$ \sum_{k=1}^{\infty} x^{2k+1} = \boldsymbol{x} + x^3 + x^5 + ... - \boldsymbol{x} = x(1 + x^2 + (x^2)^2 + ...) - x $$ 
$$ = x\left(\frac{1}{1-x^2}\right) - x = \frac{x^3}{1-x^2}$$
                $$ \sum_{k=1}^{\infty} (2k+1)x^{2k} = \frac{d}{dx}\left( \sum_{k=1}^{\infty} x^{2k+1}\right) = \frac{d}{dx}\left( \frac{x^3}{1-x^2}\right) = \boxed{\frac{3x^2-x^4}{(1-x^2)^2}}$$
And here are my steps for SBP, with a different result:
$$\sum_{k=1}^{\infty}(2k+1)x^{2k} $$
                $$ u_k = 2k+1, \triangle u_k = 2(k+1) + 1 - (2k + 1) = 2 $$
                $$ \triangle v_k = v_{k+1} - v_k = \frac{x^{2(k+1)}}{x^2-1} - \frac{x^{2k}}{x^2-1} = \frac{x^{2k}(x^2-1)}{x^2-1} = x^{2k} $$
                $$\sum_{k=1}^{\infty}(2k+1)x^{2k} = (2n+1)\left( \frac{x^{2n}}{x^2-1} \right)\bigg|_{n\rightarrow\infty} - \sum_{k=1}^{\infty} 2\left(\frac{x^{2k+2}}{x^2-1}\right)$$
                $$ = 0 - \sum_{k=1}^{\infty} 2\left(\frac{x^{2k+2}}{x^2-1}\right) \textit{ since } |x| < 1$$
                $$ = -\frac{2x^2}{x^2-1}\sum_{k=1}^{\infty} x^{2k} = -\frac{2x^2}{x^2-1}\left( \frac{x^2}{1-x^2} \right) = \frac{2x^2}{1-x^2}\left( \frac{x^2}{1-x^2} \right) = \boxed{\frac{2x^4}{(1-x^2)^2}}$$
Can anyone point out my mistake(s)?

By setting $v_k = \frac{x^{2k}}{x^2-1}$ we have $v_{k+1}-v_k = x^{2k}$, hence:
$$\begin{eqnarray*} \sum_{k=1}^{n} (2k+1)(v_{k+1}-v_{k}) &=& (2n+1)(v_{n+1}-v_1)-\sum_{k=1}^{n\color{red}{-1}}2(v_{k+1}-v_1)\\&=&\color{red}{-v_1}+(2n+1)v_{n+1}-2\sum_{k=1}^{n-1}v_{k+1}\end{eqnarray*} $$
and by letting $n\to +\infty$:
$$ \sum_{k=1}^{+\infty}(2k+1)x^{2k} = \frac{x^2}{1-x^2}+\frac{2}{1-x^2}\sum_{k\geq 1}x^{2k}=\frac{x^2(1-x^2)+2x^2}{(1-x^2)^2}.$$
By the way, the fastest method (IMHO) to compute such a series is neither differentiation nor summation by parts. If we denote with $\Delta$ the backward difference operator, $\Delta a_{n} = a_{n}-a_{n-1} $, we have:
$$ (1-x)\sum_{n\geq 0} a_n x^n = \sum_{n\geq 0} \Delta a_n x^n. $$
Assuming $a_n=2n+1$, that is a polynomial of degree $1$ in $k$, $\Delta^2 a_n=0$ for any $n\geq 2$.
It follows that:
$$ (1-x^2)^2 \sum_{k\geq 1}(2k+1)x^{2k} = c_0 + c_1 x^2 + c_2 x^4$$
where $c_0,c_1,c_2$ are easy to find.
