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Find the real number $a$ such that the sequence $a_n=1^9+2^9+...+n^9-an^{10}$ has a finite limit.

My answer is that it doesn''t exist such an a, because the first sum before the minus sign is a polynomial $P(n)$ with degree 10 and leading coefficient $1/10$ . But i am interested if you have another approach. for my proof to be complete i would need to show that the polynomial P has at least another non zero coefficient (different from the constant term). This surely is true, but how do i show it?

amarius8312
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1 Answers1

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Let's assume even less and we just 'know' that $\sum_{i=1}^ni^9$ is a polynomial. That means $\sum_{i=1}^ni^9 - an^{10}$ is also a polynomial and in order to have a finite limit for $n\to\infty$, it has to be constant ($b$), that means we would have

$$f(i):=\sum_{i=1}^ni^9 = an^{10}+b.$$

Simple calculations shows $f(1)=1, f(2)=513$, that means $f(2)-f(1)=512=a(1024-1)$, from which $a=\frac{512}{1023}$ follows and $b=\frac{511}{1023}$. But then we would have $f(3)=\frac{512\times3^{10}+511}{1023} \notin \mathbb Z$.

If the big numbers bother you, you can argue that the above formula should also hold for $n=0$, which leads to $b=0$ immediately and $f(1)$ and $f(2)$ show the contradiction with managable numbers.

Ingix
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