Given that $$f(x) = \frac{2}{4^x + 2} \forall x\in \mathbb{R}$$ Evaluate the sum $$f\left(\frac{1}{2001}\right)+f\left(\frac{2}{2001}\right)+...+f\left(\frac{2000}{2001}\right)$$
I found that $\sum_{n=1}^{2000}\left(\frac{1}{2^{\frac{2n-1}{2001}}+1}\right)$ is not an arithmetic series, nor a geometric one.
I've also failed to find a nice recurrence relation between the sum terms.
This question probably has a clever telescoping series approximation, wich I couldn't develop by my own. Can you help me with this problem?
Observe that for any $x \in \mathbb{R}$, $$f(1-x) = \frac{2}{4^{1-x} + 2} = \frac{2 \cdot 4^x}{4 + 2 \cdot 4^x} = \frac{4^x}{2 + 4^x} = 1 - f(x).$$ Hence you can evaluate the sum simply by pairing up terms.