Is there an infinite sequence of real numbers $a_1, a_2, a_3,...$ such that ${a_1}^m+{a_2}^m+a_3^m+...=m$ for every positive integer $m$?
I tried assuming that the sequence $a_1^m, a_2^m,...$ forms a geometric progression, because that is the only type of infinite series that I know how to evaluate. I know my attempt doesn't work for all integers $m$, but it does work for $m=1$:
Let $a_1=\dfrac 12$ and $a_n=a_1^n$
We have $a_1^m+a_2^m+a_3^m+...=(a_1)^m+(a_1^2)^m+(a_1^3)^m+...= \left( \dfrac 12 \right)^m+ \left(\dfrac 12 \right)^{2m}+ \left(\dfrac 12 \right)^{3m}+...=\sum_{i=1}^\infty \left( \dfrac 12 \right)^{i \cdot m}=m$
Now if we let $m=1$, we have $\sum_{i=1}^\infty \left( \dfrac 12 \right)^{i}=\dfrac {\dfrac 12}{1-\dfrac 12}=1$