1. **Problem Statement:** We have three infinite series: $$K=\sum_{n=1}^\infty \left(\frac{1}{n}\right)^a x_n, \quad L=\sum_{n=1}^\infty \left(\frac{1}{n}\right)^b x_n, \quad M=\sum_{n=1}^\infty \left(\frac{1}{n}\right)^c x_n$$ where $0 < a < c < b < 1$, $K=0$, $L=0$, and $x_n \in [-1,1]$ with possibly varying values. We want to prove that $M=0$. 2. **Key Idea:** Since $a < c < b$, the weights $\left(\frac{1}{n}\right)^a$, $\left(\frac{1}{n}\right)^c$, and $\left(\frac{1}{n}\right)^b$ satisfy $$\left(\frac{1}{n}\right)^b < \left(\frac{1}{n}\right)^c < \left(\frac{1}{n}\right)^a$$ for all $n \geq 1$ because the function $f(t) = n^{-t}$ is strictly decreasing in $t$ for fixed $n > 1$. 3. **Using the given equalities:** We know $$\sum_{n=1}^\infty \left(\frac{1}{n}\right)^a x_n = 0$$ and $$\sum_{n=1}^\infty \left(\frac{1}{n}\right)^b x_n = 0.$$ 4. **Goal:** Show that $$\sum_{n=1}^\infty \left(\frac{1}{n}\right)^c x_n = 0.$$ 5. **Approach:** Since $x_n$ are bounded in $[-1,1]$, consider the difference between the series: $$\sum_{n=1}^\infty \left(\frac{1}{n}\right)^a x_n - \sum_{n=1}^\infty \left(\frac{1}{n}\right)^b x_n = 0 - 0 = 0.$$ Rewrite this as $$\sum_{n=1}^\infty \left[\left(\frac{1}{n}\right)^a - \left(\frac{1}{n}\right)^b\right] x_n = 0.$$ 6. **Since $a < b$, for each $n$:** $$\left(\frac{1}{n}\right)^a - \left(\frac{1}{n}\right)^b > 0,$$ so the weights $w_n = \left(\frac{1}{n}\right)^a - \left(\frac{1}{n}\right)^b$ are positive. 7. **Interpretation:** The series $$\sum_{n=1}^\infty w_n x_n = 0$$ with $w_n > 0$ and $x_n \in [-1,1]$ implies that the weighted average of $x_n$ with positive weights $w_n$ is zero. 8. **Similarly, consider the weights for $M$:** Since $a < c < b$, we have $$\left(\frac{1}{n}\right)^b < \left(\frac{1}{n}\right)^c < \left(\frac{1}{n}\right)^a,$$ so $$\left(\frac{1}{n}\right)^a - \left(\frac{1}{n}\right)^c > 0, \quad \left(\frac{1}{n}\right)^c - \left(\frac{1}{n}\right)^b > 0.$$ 9. **Express $M$ as a convex combination:** There exists $\lambda \in (0,1)$ such that $$\left(\frac{1}{n}\right)^c = \lambda \left(\frac{1}{n}\right)^a + (1-\lambda) \left(\frac{1}{n}\right)^b.$$ This is because $\left(\frac{1}{n}\right)^c$ lies strictly between $\left(\frac{1}{n}\right)^a$ and $\left(\frac{1}{n}\right)^b$ for each $n$. 10. **Use linearity of series:** Then $$M = \sum_{n=1}^\infty \left(\frac{1}{n}\right)^c x_n = \sum_{n=1}^\infty \left[\lambda \left(\frac{1}{n}\right)^a + (1-\lambda) \left(\frac{1}{n}\right)^b\right] x_n = \lambda K + (1-\lambda) L = \lambda \cdot 0 + (1-\lambda) \cdot 0 = 0.$$ **Final answer:** $$\boxed{M=0}.$$