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We know the following sum is a polynomial of degree $N+1$ about $n$ where $n, N\in\mathbb N$: $$S_N(n)=\sum_{k=1}^{n}k^N=1^N+2^N+3^N+\cdots+n^N.$$

Then, I got interested in the following question:

Do there exist polynomials $F(x), G(x)$ such that $S_{-1}(n)=\frac{F(n)}{G(n)}$ for any natural number $n$?

The answer must be $NO$ (otherwise we shold know it!), but I don't know how to prove this.

Then, here is my question.

Question: Could you show me how to prove that there don't exist polynomials $F(x), G(x)$ such that $S_{-1}(n)=\frac{F(n)}{G(n)}$ for any natural number $n$?

Martin Sleziak
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mathlove
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1 Answers1

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Using the sums $S_{-1}(n)$ as estimates for the area under the graph of $1/x$ we get the inequalities $$ \int_1^n\frac1x\,dx<S_{-1}(n)<1+\int_1^n\frac1x\,dx. $$ Therefore $$ \ln n<S_{-1}(n)<1+\ln n. $$ This gives us a contradiction, when $n\to\infty$:

If $\deg F>\deg G$, then $F(n)/(n G(n))$ would be bounded away from zero for large $n$, but $\ln n/n\to0$ as $n\to\infty$, so the above estimates show that $S_{-1}(n)/n\to0$ when $n\to\infty$.

On the other hand, if $\deg F\le \deg G$, then $F(n)/G(n)$ remains bounded for large $n$ contradicting the fact that $\ln n$, and hence $S_{-1}(n)$ is unbounded.

Edit: See Gerry Myerson's answer to an earlier question for more related information.

Community
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Jyrki Lahtonen
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