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 $m$? 
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 $\sum_{i=1}^\infty \left( \dfrac 12 \right)^{i}=\dfrac {\dfrac 12}{1-\dfrac 12}=1$
 A: Such a sequence doesn't exist. In fact, there is no sequence of real numbers (be it finite or infinite) such that
$$\sum_{k} a_k^m = m\quad\text{ for }\quad m = 2, 3, 4$$
Assume the contrary, let's say there is indeed such a sequence.
Apply Cauchy-Schwarz to $(a_k)$ and $(a_k^2)$, we find:
$$9 = 3^2 = \left(\sum_{k} a_k^3\right)^2 = \left(\sum_{k} a_k\cdot a_k^2 \right)^2 \le \left(\sum_{k} a_k^2 \right)\left(\sum_{k} a_k^4 \right) = 2\cdot 4 = 8$$
which is clearly impossible.
A: Suppose we have a sequence $\{a_i\}$ with the desired property.  This will lead to a contradiction.
We will consider only even $m$, hence we may assume without loss of generality that all $a_i$ are nonnegative.  If $a_k>1$ for any $k$, then the LHS is at least $a_k^m$, which is greater than $m$ for sufficiently large $m$.  This is a contradiction, so therefore all $a_i\in [0,1]$.  But now $a_i^{m+2}\le a_i^m$ for all $i$, so the LHS is nonincreasing with $m$.  This is again a contradiction, as the RHS is increasing with $m$.
A: Summing up your idea with the modified geometric series, $a_i = q^i$:
$$
\sum_{i=0}^N q^{im} = 
\sum_{i=0}^N (q^m)^i = 
\frac{1 - (q^m)^{N+1}}{1 - q^m}
\to
\frac{1}{1 - q^m}
$$
for $N \to \infty$ if $\lvert q^m \rvert < 1$.
Additionally you want:
$$
\frac{1}{1-q^m}= m \iff \\
q^m = 1 - \frac{1}{m} = \frac{m-1}{m} \iff \\
q = \left( \frac{m-1}{m} \right)^{1/m}
$$
Example $m=2$: Then $q = ((2-1)/2)^{1/2} = 1/\sqrt{2}$ and we get
$$
\sum_{i=0}^\infty q^{im} = 
\sum_{i=0}^\infty (1/\sqrt{2})^{2i} =
\sum_{i=0}^\infty ((1/\sqrt{2})^i)^2 =
\sum_{i=0}^\infty (1/2)^i =
\frac{1}{1-1/2} = 2 = m
$$
as $\lvert 1/2 \rvert < 1$.
