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Let $a_0,\ldots, a_n$ be real numbers such that $\displaystyle \sum_{k=0}^n \frac{a_k}{k+1} = 0$. Prove that there exists a real number $x$ for which $\displaystyle \sum_{k=0}^n a_kx^k = 0$.

Attempt

We have $\dfrac{a_0}{1}+\dfrac{a_1}{2}+\cdots+\dfrac{a_n}{n+1} = 0$, so to relate that to $a_0 + a_1x+\cdots+a_nx^n$ we get a common denominator $(n+1)!$ so that $a_0(n+1)!+a_1\dfrac{(n+1)!}{2}+\cdots+a_n\dfrac{(n+1)!}{n+1}=0$. Then how do I transform this so I have $x,x^2,\ldots,x^k$ multiplying the terms?

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Let $f(x)= \sum a_kx^k$. Then $f(x)=g'(x)$ where $g(x)=\sum \frac{a_k}{k+1}x^{k+1}$. Also, $g(0)=g(1)=0$ so by Rolle's theorem $g'(a)=0$ for some $a \in (0,1)$.

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  • $\begingroup$ Isn't $g(x) = 0$ for all $x$? $\endgroup$
    – user19405892
    Mar 14, 2016 at 22:37
  • $\begingroup$ Here's a real example $a_0=1$, $a_2=-3$, and $a_i=0$ for all other $i$. You can see that $a_0+1/3(a_2)=0$. In this case $f(x)=1-3x^2$, $g(x)=x-x^3$. $\endgroup$
    – Noah Olander
    Mar 14, 2016 at 22:49
  • $\begingroup$ Is it possible the sequence can also be infinite? $\endgroup$
    – user19405892
    Mar 14, 2016 at 23:58
  • $\begingroup$ The same argument will work provided all the sums make sense... that is, if the series $\sum a_kx^k$ has radius of convergence greater than one. $\endgroup$
    – Noah Olander
    Mar 15, 2016 at 0:02
  • $\begingroup$ Why do we have to worry about that and not just use the same argument? $\endgroup$
    – user19405892
    Mar 15, 2016 at 0:03

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