Mathematical Reflections 2012, Issue 6 - Problem U247

Problem:
Let $a$ be a real number greater than $1$. Evaluate
$$\dfrac{1}{a^2-a+1}-\dfrac{2a}{a^4-a^2+1}+\dfrac{4a^3}{a^8-a^4+1}-\dfrac{8a^7}{a^{16}-a^{8}+1}+\ldots.$$

Proposed by Titu Andreescu.

Solution:
Let $$S_n=\sum_{k=1}^n (-1)^{k-1}\dfrac{2^{k-1}a^{2^{k-1}-1}}{a^{2^k}-a^{2^{k-1}}+1}.$$ Since
$$\begin{array}{rcr} \dfrac{1}{a^2-a+1}-\dfrac{1}{a^2+a+1}&=&\dfrac{2a}{a^4+a^2+1} \\  -\dfrac{2a}{a^4-a^2+1}+\dfrac{2a}{a^4+a^2+1}&=&-\dfrac{4a^3}{a^8+a^4+1} \\ \vdots & \vdots & \vdots \\ (-1)^{n-1} \dfrac{2^{n-1}a^{2^{n-1}-1}}{a^{2^n}-a^{2^{n-1}}+1}+(-1)^n \dfrac{2^{n-1}a^{2^{n-1}-1}}{a^{2^n}+a^{2^{n-1}}+1} & = & (-1)^{n+1} \dfrac{2^na^{2^n-1}}{a^{2^{n+1}}+a^{2^n}+1}. \end{array}$$
summing up the two columns, we get $$S_n-\dfrac{1}{a^2+a+1}=(-1)^{n+1} \dfrac{2^na^{2^n-1}}{a^{2^{n+1}}+a^{2^n}+1}.$$
Now, $$0 \leq \left|(-1)^{n+1}\dfrac{2^na^{2^n-1}}{a^{2^{n+1}}+a^{2^n}+1}\right| \leq \dfrac{2^na^{2^n}}{a^{2^{n+1}}}=\dfrac{2^n}{a^{2^n}}$$ and  $\dfrac{2^n}{a^{2^n}} \to 0$ as $n \to \infty$. So,
$$\sum_{k=1}^\infty (-1)^{k-1}\dfrac{2^{k-1}a^{2^{k-1}-1}}{a^{2^k}-a^{2^{k-1}}+1}=\lim_{n \to \infty} S_n = \dfrac{1}{a^2+a+1}.$$