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Let $(a_n),(b_n)$ two summable sequences. Show that if $\sum_{n=0}^{\infty}a_n^k=\sum_{n=0}^{\infty}b_n^k$ for all $k\in \Bbb N^*$ then $(a_n) = (b_n)$ more or less a permutation

What I tried: to deal with the "more or less a permutation", I decided to re-index the sequences by:

  • decreasing modulus as first
  • If there is equality, decreasing real part
  • if still equal, decreasing imaginary part

(Assuming that the sequences are complexes).

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  • $\begingroup$ Which values does $k$ assume? $\endgroup$ – Lorenzo Quarisa Jun 13 '18 at 8:49
  • $\begingroup$ $\Bbb N^*$. (Thanks, I forgot to say it) $\endgroup$ – MiKiDe Jun 13 '18 at 8:57
  • $\begingroup$ This is an interesting claim. Where did you get this? $\endgroup$ – Aryabhata Jun 13 '18 at 22:44
  • $\begingroup$ Any constraints on $a_n$ and $b_n$ other than summable? We may have $b_n=a_{n+1}$. $\endgroup$ – i707107 Jun 13 '18 at 23:33
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    $\begingroup$ We must assume that $a_n \neq 0 \neq b_n$ for all $n$. Or at least that both sequences contain the same amount of $0$s, for inserting zeros doesn't change any of the sums. // If you know a little complex analysis, consider the function $$f(z) = \sum_{n = 0}^{\infty} \frac{a_nz}{1 - a_nz}$$ and the analogous function for the $b_n$. Conclude that they are equal. $\endgroup$ – Daniel Fischer Jun 16 '18 at 13:00
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Here's one way to approach this :

$(a_n)$ is summable hence converges to $0$, hence the $\sup$ of its modulus is attained. Up to a permutation, it's $a_0$. Divide both sequences by $a_0$ , you still have the same inequality.

Now by dominated convergence (or some $\epsilon-\delta$-work, which would also provide this) $\displaystyle\sum_{n, |a_n| < |a_0|} (\frac{a_n}{a_0})^k$ converges to $0$ as $k\to \infty$. Similarly for $b$ by replacing the indexing by "$n, |b_n|<|a_0|$".

Now since the sequence is summable, there is only a finite set $I_a$ (resp. $I_b$) of indices $n$ such that $|a_n|=|a_0|$(resp. $|b_n|=|b_0|$).Write $I_a = J_a\sqcup E_a$ (resp. $I_b = J_b\sqcup E_b$) where $J_a$ is those indices $n$ for which $a_n\neq a_0$, and $E_a$ the other ones (resp. blabla).

Then, writing $f(k)$ to be the $k$th sum we get $f(k) = v_k + |E_a|+ \displaystyle\sum_{n\in J_a}(\frac{a_n}{a_0})^k$, where $v_k\to 0$.

$J_a$ is finite, and for $n\in J_a$, $(\frac{a_n}{a_0})^k$ Cesaro-converges to $0$ as $k\to \infty$, we get by the Cesaro convergence theorem that $f(k)$ Cesaro-converges to $|E_a|$.

But if we do the same thing with $b$ we get that $f(k)$ Cesaro-congerges to $|E_b|$.

Hence $|E_b|=|E_a|$.

We can thus remove all the $a_0$ terms of $a$ and $b$ and do the same thing again for all terms of modulus $|a_0|$, until we are left with only terms of modulus $<|a_0|$ in $a$ (and thus in $b$ !)

We do this again, at each step we look at which of the sequences has the biggest $\sup$ of modulus, do what we just did and find that actually both had the same $\sup$ of modulus with the same number of indices having said modulus and more precisely the same number of indices having a given modulus and argument.

Since the modulus of each sequence tends to zero, we are guaranteed that this process will "reach" (this is the bit that's annoying to formalize) every integer, and thus the connection we will have made from $a$ to $b$ will be a permutation of $\mathbb{N}$

(Actually with the reordering that you proposed, this last step is not annoying because everything works in the order you gave, and we actually get that the sequences are equal, so without the reduction, equal up to a permutation)

EDIT : I don't know if it's standard so I'll clarify a bit : a sequence $(u_n)$ Cesaro-converges to $l$ if $\frac{1}{n+1}\displaystyle\sum_{k=0}^n u_k$ converges to $l$. The Cesaro convergence theorem states that if $u_n$ converges to $l$, then it Cesaro-converges to $l$. This is easy to prove and is a good exercise for a first course in calculus.

One then easily checks that if $z$ is a complex of modulus $1$ that is not equal to $1$, then $(z^n)$ Cesaro-converges to $0$ (this is because the sums are bounded)

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Starting with $$ \sum\limits_{0\, \le \,n} {a_{\,n} ^m } = \sum\limits_{0\, \le \,n} {b_{\,n} ^m } \quad \left| {\;\forall m \in \mathbb N} \right. $$

a) We can assume that $\left| {\,a_{\,k} \,} \right| < 1\; \wedge \;\left| {\,b_{\,j} \,} \right| < 1$, because ,otherwise, we can always divide both sides by the maximum of the moduli in $a$ and $b$ terms, and in case augmented by an additive or multiplicative constant.

b) We can then multiply by $x^m$ and sum on $m$ $$ \sum\limits_{0\, \le \,n} {\sum\limits_{0\, \le \,m} {a_{\,n} ^m x^{\,m} } } = \sum\limits_{0\, \le \,n} {\sum\limits_{0\, \le \,m} {b_{\,n} ^m x^{\,m} } } $$ being assured that the double sum will be convergent (and absolutely) for $|x|<1$.

c) We obtain $$ f(x) = \sum\limits_{0\, \le \,n} {\sum\limits_{0\, \le \,m} {a_{\,n} ^m x^{\,m} } } = \sum\limits_{0\, \le \,n} {{1 \over {1 - a_{\,n} x}}} = \sum\limits_{0\, \le \,n} {{{ - 1/a_{\,n} } \over {x - 1/a_{\,n} }}} $$ but, by the Mittag-Leffler's theorem this is the partial-fractions expansion of a meromorphic function with poles in $1/a_n$, which is unique apart from the ordering of the summands,
which in fact are normally ordered as you said.

Therefore the hypothesis is demonstrated.

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