# 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 $\sum_{i=1}^\infty \left( \dfrac 12 \right)^{i}=\dfrac {\dfrac 12}{1-\dfrac 12}=1$

• I couldn't understand the step $\sum_1^\infty \left(\frac{1}{2}\right)^{i\cdot m}=m$... should'nt $\sum_1^\infty \left(\frac{1}{2}\right)^{i\cdot m}=\frac{1}{2^m-1}$?? – SchrodingersCat May 25 '16 at 4:12
• @SchrodingersCat If we let $m=1$, we just have $\sum_{1}^{\infty} \left( \dfrac 12 \right) ^m$ – Ovi May 25 '16 at 4:15
• @mvw Oops that's a mistake then, but I believe the formula $S=\dfrac {a}{1-r}$, for summing my series still works, right? – Ovi May 25 '16 at 4:17
• @Ovi Then it holds only for $m=1$ and not for $m \ge 2$. You should mention it. – SchrodingersCat May 25 '16 at 4:19
• @SchrodingersCat I did mention it in the original, unedited post, I said "I know my attempt doesn't work for all integers $m$, but it does work for $m=1$". – Ovi May 25 '16 at 4:23

## 3 Answers

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 $m$.

• It's worth noting that this argument extends to the complex numbers, though takes a lot more work; there, we take all the summands of greatest absolute value, note that their sum acts exponentially for some sequence of $n$. But the summands of lesser absolute value vanish too quickly to matter. – Milo Brandt May 25 '16 at 4:25

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.

• Impressive!$\,$ – lisyarus May 25 '16 at 6:49

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 $\lvert 1/2 \rvert < 1$.

• For every +ve integer $m$, one could find such a sequence. That the same unique sequence works for all $m$ does not seem to be the case. – vnd May 25 '16 at 4:46
• I see. I read the task differently. Thanks.. Here we have $q =q(m)$, not a general $q$ working for all $m$. – mvw May 25 '16 at 4:48