TITLE: Proving there don't exist $F(x), G(x)$ such that $1^{-1}+2^{-1}+3^{-1}+\cdots+n^{-1}={F(n)}/{G(n)}$
QUESTION [6 upvotes]: 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 $n$?

REPLY [9 votes]: 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.