I want to compute a simple upper bound for
$$S = \sum_{k=1}^{T-1} (k x^{k-1})^2$$
that depends on both $x \in [-1,1]$ and $T$.
I can compute
$$S = \frac{(-2 T^2 + 2 T + 1) x^{2 T + 2} + T^2 x^{2 T} + (T - 1)^2 x^{2 T + 4} - x^4 - x^2}{x^2 (x^2 - 1)^3}.$$
But can I compute an upper bound on $S$ that is simpler? For example, a bound that we can see immediately that $S(0) = 0$ and $S(1)$ is bounded. An obvious upper bound for this sum is when $x = 1$, which is $\sum_{k=1}^{T-1} k^2 = T(2 T^2 - 3T +1)/6$, but I'm looking for an upper bound that depends on $x$. Any suggestions?
What about the Cauchy-Schwarz inequality? It decouples the summations of $k^2$ and $x^{2(k-1)}$ and hence gives a product bound: $$ S^2 = (\sum_{k=1}^{T-1} k^2 x^{2(k-1)})^2 \le \sum_{k=1}^{T-1} k^4 \sum_{k=1}^{T-1} x^{4(k-1)}\\ = \frac{T (T - 1) (2 T - 1) (3 T^2 - 3 T - 1)}{30} \cdot \frac{x^4 - x^{4 T}}{x^4 - x^8} $$ which gives $S(x=0) = 0$ and $S^2(x=1) \le \frac{T (T - 1)^2 (2 T - 1) (3 T^2 - 3 T - 1)}{30} $.