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Suppose we know that $$\sum_{n=0}^{k}(-1)^na_n=0$$ for even $k$, and that $\{a_n\}$ is decreasing after $n/2$. What conditions can we set on $a_n$ to make $$\sum_{n=0}^k(-1)^na_n(-n)$$ either positive or negative?

Stahl
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user81654
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  • What does "decreasing after $n/2$" mean? Also, welcome, and please note that we like to see your attempt at a solution in order to help most efficiently. – not all wrong Jun 10 '13 at 00:55
  • It seems that $a_{2n}=a_{2n-1}$, at least for every $n\geqslant2$. Is this intended? – Did Jun 10 '13 at 00:57
  • sorry, by decreasing after $n/2$ I meant decreasing after $k/2$. – user81654 Jun 10 '13 at 02:25
  • Now, what I'm figuring, is that ${a_n}}$ can't always be decreasing (for every $n$), or it would not equal zero. So we must have that $a_n<a_{n+1}$ for some some $n<k/2$. – user81654 Jun 10 '13 at 02:30

2 Answers2

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I reckon you meant $\{a_n\}$ is decreasing after $\frac{k}{2}$.

In that case, let $\displaystyle a_n=e^{-\left(n-\left(\frac{k+1}{2}\right)\right)^2}$, then $\displaystyle \sum_{n=0}^{k}(-1)^na_n\approx0$ and $\displaystyle \sum_{n=0}^k(-1)^na_n(-n)$ is negative when $\frac{k}{2}$ is odd and positive when $\frac{k}{2}$ is even.

Maazul
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Ok, it seems to me that the original sum is $$S_1:=a_0-a_1+a_2-....\pm a_k=0$$ Our second sum is therefore: $$S_2:=a_1-2a_2+3a_3-4a_4+...\pm ka_k$$ I guess that if we can find an $M$ such that $a_{i+1}/a_i=M(i+1)$, and we define $M(1)=1$, then we can say that $$S_2=a_1-2\Big( M(2)a_1 \Big)+3\Big( M(3)M(2)a_1\Big)-...\pm k\Big(a_1 \prod_{i=1}^{k} M(i) \Big)$$ $$=a_1 \sum_{j=1}^{k}(-1)^{j+1}j\Big( \prod_{i=1}^{j}M(i) \Big)$$ Now, we can safely say that if $\{b_n\}$ is an increasing sequence, then $$\sum_{i=1}^k(-1)^{i+1}b_n n$$ is positive if $k$ is odd and negative otherwise. Therefore, if we have that $$V(j)=\prod_{i=1}^{j}M(i)$$ is an inreasing function of $j$, then we can govern and control the positivity of $S_2$ as is evident in the example given by Maazul. Of course, $V(j)$ will only be increasing if $\{a_n\}$ is increasing. So this is all pretty useless.

user81654
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