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Let $S_n$ be the harmonic series $$S_n = \sum_1^{n} \frac{1}{n}$$ Is there a polynomial $p(x)$ with real coefficients so that $p(n) = S_n$ for all integers $n$?

user830529
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    No, because $\sum_{k=1}^{n} \frac{1}{k}\sim \ln(n)$. See https://en.wikipedia.org/wiki/Harmonic_number – Robert Z Oct 13 '20 at 05:58
  • Suppose $p$ is such a polynomial. Then $|p(n) | \leq n$. This implies the polynomial $q(x)$ where $q(x)=\frac {p(x)-p(0)} x$ (for $ x\neq 0$ and $p'(0)$ for $x=0$) is bounded along the sequence ${1,2,3...}$. This implies that $q$ is a constant, say $c$. So $p(x)=p(0)+cx$ which clearly cannot satisfy the condition $p(n)=S_n$ for all $n$. – Kavi Rama Murthy Oct 13 '20 at 05:59

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Assume by contradiction that there is such a polynomial $P(X)$ of degree $k$.

Define $$Q(X):=P(X+1)-P(X)$$

Then $Q(x)$ is a polynomial of degree $k-1$ and satisfies $$Q(n)=S_{n+1}-S_n=\frac{1}{n+1}$$

Now you can reach a contradiction one of the following two ways:

  • Since $\lim_{n \to \infty} Q(n)=0$ we get that $Q =0$ which contradicts the fact that $Q(n)=\frac{1}{n+1}$
  • If you do not want to use limits, observe that $$nQ(n)+Q(n)=1$$ Therefore, the polynomials $1$ and $XQ(X)+X$ agree everywhere, meaning $$(X+1)Q(X)=1$$ which is not possible since the LHS is either 0 or has degree at least 1.
N. S.
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I think that it should read $S_n = \sum_{k=1}^{n} \frac{1}{k}.$

Suppose that there is a polynomial $p$ of degree $m$ such that $p(n)=S_n$ for all $n$. It is easy to see that $m \ge 2.$

Let $p(x)=a_mx^m+a_{m-1}x^{m-1}+....+a_1+a_0$ , where $a_m \ne 0.$ From $p(n)=S_n$ we get

$$a_m + \frac{a_{m-1}}{n}+....+\frac{a_0}{n^m}= \frac{S_n}{n^m}.$$

But $\frac{S_n}{n^m} \to 0$ as $ n \to \infty$, since $m \ge 2.$ This gives

$$a_m=0,$$

a contradiction.

Fred
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There isn't a polynomial , but I have an exact formula for $S_n$: $$S_n=\gamma+\log n+\frac{1}{2n}-\sum_{k\geq 2}\frac{A_k}{n(n+1)(n+2)...(n+k-1)}$$ where $$A_k=\frac{1}{k}\int_{1}^{0}x(1-x)(2-x)(3-x)...(k-1-x)dx,\text{and}$$ $\gamma$ is the Euler-mascheroni constant.
Some values of $A_k$: $$A_2=\frac{1}{12}$$ $$A_3=\frac{1}{12}$$ $$A_4=\frac{19}{120}$$ $$A_5=\frac{9}{20}$$ This information is from the book table of integrals, series and products by I.S. gradshteyn I.M. ryzhik.