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let sequence $\{a_{n}\}$,such $$a_{n}=\displaystyle\sum_{i=1}^{n}i^k=1+2^k+3^k+\cdots+n^k$$where $k$ is real numbers,

show that:

there is exsit polynomial $f(x)$, such for any $n\in Z^{+}$,always have $f(n)=a_{n}\Longleftrightarrow k\in Z^{+}$

my try: if $k$ is positive integer,then $$a_{n}=1+2^k+3^k+\cdots+n^k$$ is postive integers,and is polynomial with degree $n$,so let $$f(x)=x^k+(x-1)^k+\cdots+2^k+1^k$$

But if for any $n\in Z^{+}$,then exsit the polynomial $f(x)$ have $f(n)=1^k+2^k+\cdots+n^k$

How prove $k$ is positive integers?

Thank you very much!enter image description here

please @ Ivan Loh and so on help.because I think This is interesing problem!

Community
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math110
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    Interesting problem +1 –  Dec 02 '13 at 06:21
  • Your attempt does not work: What is the polynomial that you mean, when $k=1$? $1+2+\dots+x$ does not make sense if $x$ in not an integer, while if $x$ is an integer, then it is not clear it is a polynomial, which is precisely what you need to show. You may want to look up Bernoulli polynomials. (This still leaves pending the issue that $f$ is polynomial only when $k$ is a positive integer.) – Andrés E. Caicedo Dec 02 '13 at 07:37
  • @AndresCaicedo,Hello,my problem is exsit polynomial $f(x)$,the $f(x)$ ：http://en.wikipedia.org/wiki/Polynomial – math110 Dec 02 '13 at 08:04

1 Answers1

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Suppose $k\neq -1$. Note that $\frac{a_n}{n^{k+1}}$ is a Riemann sum associated to $g(t)=t^k$ for $t\in [0,1]$ (thanks to Daniel Fischer for pointing this to me). So it converges to $\int_{0}^1 g(t)dt=\frac{1}{k+1}$.

So we have $a_n \sim \frac{n^{k+1}}{k+1}$ when $n\to\infty$. If $a_n$ is polynomial in $n$, this implies that $k+1$ is a nonnegative integer, and we are done.

Finally, if $k=-1$ we have the harmonic series which is well-known to be asymptotically equivalent to ${\sf log}(n)$, which increases slower than any polynomial.

Ewan Delanoy
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  • why $a_{n}$ is polynomial in $n$,that implies that $k+1$ is a nonngative integer? – math110 Dec 02 '13 at 15:00
  • Because if $a_n$ is polynomial with degree $d$, say $a_n=c_0+c_1n+c_2n^2+\ldots +c_dn^d$, then $a_n \sim c_nn^d $ when $n\to \infty$. So we must have $c_dn^d \sim \frac{n^(k+1)}{k+1}$ when $n\to \infty$, and this forces $d=k+1$. – Ewan Delanoy Dec 02 '13 at 15:05