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I wish to know if there are real sequences $(a_k)$, $(b_k)$ (and if there are, how to construct such sequences) such that:

$b_k<0$ for each $k \in \mathbb{N}$ with $\lim\limits_{k \rightarrow \infty} b_k=-\infty,$

$$\sum_{k=1}^\infty |a_k| |b_k|^n< \infty, \space\forall n\in\mathbb{N}\cup\{0\}$$

$$\sum_{k=1}^\infty a_k b_k^n=1, \space \forall n\in\mathbb{N}\cup\{0\}$$

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what about $b_k = -k$ with $a_1 = -1$ and $a_k =0$ for $k \ge 2$ ? –  mercio Jul 27 '12 at 11:39
1  
@mercio: Then for every even positive $n$, the sum is $-1$ instead of $1$. –  Did Jul 27 '12 at 12:48

3 Answers 3

up vote 3 down vote accepted

For any unbounded sequence $(b_k)$ and any sequence $(\lambda_n)$, you can find a sequence $(a_k)$ such that for any $n$, $\sum_{k=1}^\infty a_k b_k^n$ converges absolutely to $\lambda_n$ :

Suppose you have constructed $a_k$ up to some index $k_0$, such that for all $m < n$, $\sum_{k=1}^{k_0} a_k b_k^m = \lambda_m$. The goal now is to extend this up to some index $k_1$ such that for all $m \le n$, $\sum_{k=1}^{k_1} a_k b_k^m = \lambda_m$, with "arbitrarily small coefficients".

This means we have to add $y = \lambda_n - \sum_{k=1}^{k_0} a_k b_k^n$ to the partial sum of the $n$th series, and $0$ to the first $n-1$ series. In order to do that, look at the next $n-1$ distincts values of $b_k$ (for $k > k_0$), say they are $c_1,c_2, \ldots c_{n-1}$, look at the Vandermonde matrix

$$ M(X) = \begin{pmatrix} c_1^1 & \ldots & c_{n-1}^1 & X^1 \\ \vdots & & \vdots & \vdots \\ c_1^n & \ldots & c_{n-1}^n & X^n \end{pmatrix}$$

and solve for the vector $A(X)$ in the equation $M(X) A(X) = (0,0,\ldots,0,y)$. We get that $A(X)$ has to be $y$ times the $n$th column of $M(X)^{-1}$. Thus $A(X)$ is $y/det(M(X))$ times the transpose of the $n$th row of the comatrix of $M(X)$. If you look carefully at this row, each entry there is a polynomial in $X$ of degree $n-1$, except the last one which is a constant. Furthermore, the determinant of $M(X)$ is of degree $n$, so the entries of the vector $A(X)$ are rational fractions of degree $-1$ and $-n$.

In particular, for $X$ large enough, $\sum_{1 \le i,j \le n-1} |A_i(X)c_i^j| < 2^{-n}$ Thus, if you pick $k_1$ such that $b_{k_1}$ is large enough, you will get from $A(b_{k_1})$ the coefficients to put in front of $c_1^n, c_2^n, \ldots, c_{n-1}^n, b_{k_1}^n$ such that all the partial sums are what we want, and the absolute contribution to what we added to each partial sum is less than $2^{-n-1}$.

Now, repeat this procedure for all $n$, and you get a suitable sequence $(a_k)$.

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Let $(b_n)$ be any negative sequence diverging to $-\infty$, and suppose such a sequence $(a_k)$ exists.

Let $K \in \mathbb{N}$ be such that $b_k < -2$ for all $k \ge K$. Then

$$\sum_{k=1}^{\infty} \left| a_k \right| \left| b_k \right|^n \ge \sum_{k=K}^{\infty} \left| a_k \right| \cdot 2^n = 2^n \sum_{k=K}^{\infty} \left| a_k \right|$$

So for all $k \ge K$ we require $a_k = 0$. So then

$$\sum_{k=1}^{\infty}a_kb_k^n = \sum_{k=1}^{K-1}a_kb_k^n = 1$$

And so for all $n \in \mathbb{N}$ we have

$$\sum_{k=1}^{K-1} a_k(b_k^n - b_k^{n-1}) = 0$$

So in $\mathbb{R}^{K-1}$ the vector $(a_k)_1^{K-1}$ is linearly independent from the vectors $(b_k^{n-1}(b_k - 1))_1^{K-1}$ for each $n \in \mathbb{N}$. Since $b_k \ne 1$ and $(a_k) \ne 0$, this must mean that the $(b_k^{n-1}(b_k-1))$ all lie in the $(K-2)$-hyperplane orthogonal to $(a_k)$. But unless each $b_k=-1$, these span a $(K-1)$-dimensional subspace (i.e. the whole of $\mathbb{R}^{K-1}$), so we must have $b_k=-1$ for each $k$. Therefore

$$\sum_{k=1}^{K-1} (-1)^na_k = 1$$

for all $n \in \mathbb{N}$. This is a contradiction (multiply through by $-1$).

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I don't think the bit where you use linear independence works: the $(b_k^{n-1}(b_k-1))_1^{K-1}$ are all orthogonal to $(a_k)_1^{K-1}$ but we don't know that they're linearly independent with eachother? –  Tom Hutchcroft Jul 27 '12 at 13:46
    
I didn't know you were on MSE @Tom! And yes, good point, I'll see if I can find a way round it. I might not be able to. –  Clive Newstead Jul 27 '12 at 13:48
    
I think I've thought of how to do it –  Tom Hutchcroft Jul 27 '12 at 13:50
    
@TomHutchcroft: I think I fixed it by referring to the hyperplane orthogonal to $(a_k)$ rather than the linear independence of the vectors. Give me a moment to go over the details again in my head. –  Clive Newstead Jul 27 '12 at 13:53
    
I've got a proof that doesn't use any linear algebraic arguments, I guess I'll post it –  Tom Hutchcroft Jul 27 '12 at 14:08

Let $(b_n)$ be a negative sequence diverging to $-\infty$, and suppose such a sequence $(a_k)$ exists.

Let $K \in \mathbb{N}$ be such that $(b_k) < -2$ for all $k \geq K$. Then $\sum_{1}^{\infty}\mid a_k\mid \mid b_k \mid ^n \geq \sum_K^\infty\mid a_k\mid \mid b_k \mid ^n \geq (2)^n\sum_K^\infty\mid a_k\mid $

So for all $k \geq K$ we have $a_k=0$. (So far copying Clive!)

Now write $\sum_{1}^{\infty}\mid a_k\mid \mid b_k \mid ^n = \sum_{1}^{K-1}\mid a_k\mid \mid b_k \mid ^n= \sum_{b_k>-1}^{k < K}\mid a_k\mid \mid b_k \mid ^n + \sum_{b_k=-1}^{k < K}\mid a_k\mid \mid b_k \mid ^n +\sum_{b_k<-1}^{k < K}\mid a_k\mid \mid b_k \mid ^n$

Similarly to before, we have $\sum_{1}^{\infty}\mid a_k\mid \mid b_k \mid ^n \geq \sum_{b_k<-1}^{k < K}\mid a_k\mid \mid b_k \mid ^n \geq \min_{k<K, b_k<-1}(\mid b_k \mid) ^n \sum_{b_k<-1}^{k < K}\mid a_k\mid $ so we must have that $a_k = 0$ whenever $b_k <-1$.

Now we have $\sum_{1}^{\infty} a_k b_k ^n = \sum_{b_k>-1}^{k < K} a_k b_k ^n + (-1)^n\sum_{b_k=-1}^{k < K} a_k =1$ and taking the limit as $n$ tends to $\infty$ we have that $(-1)^n\sum_{b_k=-1}^{k < K} a_k$ tends to 1, a contradiction.

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your argument in order to show that $a_k = 0$ for $k \ge K$ supposes that there is a bound on $\sum |a_k b_k^n|$ that doesn't depend on $n$. The question only says that there is a bound for each $n$, if I read it correctly. –  mercio Jul 27 '12 at 15:02
    
Oops! You're right. –  Tom Hutchcroft Jul 27 '12 at 15:16

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