If $a$ is transcendental, then is it necessarily true that $1^a+2^a+...+n^a$ is also transcendental for n>1?
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9$\begingroup$ It is known that $\alpha = \log_2(3)$ is transcendental. Now $1^\alpha + 2^\alpha = 1 + 2^{\log_2(3)} = 1 + 3 = 4$ is rational. $\endgroup$Stanley Yao Xiao– Stanley Yao Xiao ♦2025-11-20 06:00:00 +00:00Commented Nov 20 at 6:00
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1$\begingroup$ I suspect that it is unknown, whether $1+2^{\pi}+3^{\pi}$ is transcendental. $\endgroup$Gerry Myerson– Gerry Myerson2025-11-20 06:04:49 +00:00Commented Nov 20 at 6:04
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1 Answer
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The answer is definitively no. As Stanley said in a comment, a counterexemple is straightorward for $n=2$. However, even if $n>2$, the unique real solution $x$ of $$ \sum_{k=1}^n k^x=2 $$ is transcendental.
Assume that this solution $x$ is algebraic. If $x$ is rational, a routine monotonicity/inequality argument excludes any nonzero rational $x$ giving $\sum_{k=1}^n k^x=2$.
If $x$ is algebraic and irrational, then by the Gelfond–Schneider theorem every number $k^x=e^{x\ln k}$ with integer $k\ge2$ is transcendental. Thus the terms $2^x,3^x,\dots,n^x$ are all transcendental, and we have the algebraic linear relation $$ 2^x+3^x+\cdots+n^x=1. $$
It remains to rule out the possibility that finitely many such transcendental numbers, each of the form $k^x$ with the same algebraic irrational exponent $x$, could sum to the rational number $1$. This is exactly the kind of relation excluded by deeper theorems in transcendence theory: by results going back to Baker (linear forms in logarithms) one obtains linear-independence statements that imply that the numbers $e^{x\ln k}=k^x$ (for distinct integer bases $k\ge2$) are linearly independent over the field of algebraic numbers when $x$ is an algebraic irrational. Hence no nontrivial algebraic linear relation with algebraic coefficients can hold among them; in particular they cannot sum to the rational number $1$. That contradiction shows $x$ cannot be algebraic irrational.
