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Let $\sum a_{n}$ be an absolutely convergent series such that $$\sum a_{kn}=0$$ for all $k\geq 1$. Help me prove that $a_{n}=0$ for all $n$.

Thank you!

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    $\begingroup$ What's $a_{kn}$? $\endgroup$ – knucklebumpler Apr 4 '11 at 0:52
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    $\begingroup$ Please don't post in the imperative; if you are having trouble, the best course of action is to say why or where you are having trouble, and what you have tried or succeeded in doing. $\endgroup$ – Arturo Magidin Apr 4 '11 at 0:52
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    $\begingroup$ If $a_{kn}$ is the subsequence of terms with index multiple of $k$, what happens if you take $k=1$? $\endgroup$ – Arturo Magidin Apr 4 '11 at 0:53
  • $\begingroup$ @Arturo: I edited the question to make it more polite. But this goes right back to the meta thread about editing... Was this change ok or too much? $\endgroup$ – Eric Naslund Apr 4 '11 at 1:15
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    $\begingroup$ By the way, this is a nice problem. $\endgroup$ – Pete L. Clark Apr 4 '11 at 4:51
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I assume that your sequence starts at $n=1$

Hint: Notice that we can isolate each term as $$a_k= \sum_{m=1}^\infty \left( \mu(m) \sum_{n=1}^\infty a_{knm} \right)$$ where $\mu$ is the Möbius mu function. (and the index $a_{mnk}$ is the product of variables $k,m,n$) Careful, the above sum on the right hand side might not be absolutely convergent, so an argument with partial sums is needed to show why it is equal to $a_k$.

Hope that helps,

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  • $\begingroup$ sorry for mistakes ,I 've just found this website @Eric Naslund : although i'm a beginner in series i will try with your hint thank you very much $\endgroup$ – student Apr 4 '11 at 1:59
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    $\begingroup$ What a nice answer. :-) $\endgroup$ – Andrés E. Caicedo Apr 4 '11 at 2:54
  • $\begingroup$ I have struggled with your hint in vain , could you please post your full proof , thank you very much:) $\endgroup$ – student Apr 4 '11 at 21:56
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Here's a probabilistic approach.

Define two finite measures $\mu$ and $\nu$ on $\mathbb{Z}_+$ by $\mu(\lbrace n\rbrace)=a_n^+$ and $\nu(\lbrace n\rbrace)={a_n}^-$; the positive and negative parts of $a_n$. The zero sums in the problem mean that $\mu(k\mathbb{Z}_+)=\nu(k\mathbb{Z}_+)$ for every $k\geq 1$.

The collection $\lbrace k\mathbb{Z}_+ : k\geq 1\rbrace$ is a $\pi$-system (closed under finite intersections) and generates the discrete $\sigma$-field on $\mathbb{Z}_+$. The standard application${}^*$ of Dynkin's $\pi,\lambda$ systems shows that $\mu=\nu$. In particular,
$$\mu(\lbrace n\rbrace)=a_n^+ = {a_n}^-=\nu(\lbrace n\rbrace)$$ for all $n$, that is, $a_n=0$ for all $n\geq 1$.

${}^*$ For example, Lemma 1.17 on page 9 of Kallenberg's Foundations of Modern Probability.

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Here's a suggestion.. try "sieving" to first show $a_1 = 0$. You first subtract off $\sum a_{2n} = 0$, and get that the sum of $a_n$ over $n$ odd is zero. Then subtract off $\sum a_{3n}$ and you get the resulting series is zero. But now you've subtracted off each $a_{6n}$ twice, so you add back $\sum a_{6n}$. Thus you've now removed all $a_{2n}$ and $a_{3n}$ from your series. You can then proceed with removing all $a_{5n}$, then adding back all $a_{10n}$ and all $a_{15n}$, then subtracting off all $a_{30n}$ to remove all duplicates. The end result is that you've removed all $a_{2n}$, $a_{3n}$, and $a_{5n}$. If you keep doing this, you'll end out removing everything except $a_1$, which therefore is zero.

Then repeat the process to show $a_2 = 0$, first removing off $a_{3n}$, then all $a_{4n}$ and so on. Keep going in this way, until all $a_n$ are shown to be zero.

The only issue I see here is making sure you can always get rid of the duplicates, so this is why I'm just calling this as a suggestion... but it sounds plausible to me.

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  • $\begingroup$ The Problem here is that you have to remove things "an infinity" of times, which makes your move harder. Notice that this is exactly the point of Eric Naslund's hint, i.e. the Mobius mu fonction, recall it is worth $(-1)^{w(n)}$ where $w(n)$ is the number of distinct primes in the factorization of $n$ if they have no multiplicity and $0$ otherwise. Therefore, the part where you "take 2, 3, add 6", he does it with $\mu(2) = -1$, $\mu(3) = -1$, $\mu(6) = 1$. Hence what you're trying to say is that Eric's identity's your first guess, but I must say, it's cool you detailed it. $\endgroup$ – Patrick Da Silva May 13 '11 at 5:54
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Let $A(p_1, \dots, p_m)$ be the set of all integers that are divisible by the first $m$ primes $p_1, \dots, p_m$.

I claim that $$S(m) = \sum_{n \in N \setminus A(p_1, \dots, p_m)} x_n = 0.$$

If this is true, then $$S(m) = x_1 + \sum_{n \in A} x_n = 0$$ where $A \subseteq \{p_{m+1}, p_{m+1}+1, \dots\}$. That is, $$|x_1| = \left| \sum_{x \in A} x_n \right| \leq \sum_{n \geq p_m}|x_n|.$$ Because $\sum |x_n|$ converges, if we let $m \to \infty$, we get $x_1=0$.

Next we realize that the sequence $x_{kn}$ satisfies the same conditions as $x_n$. Hence the same reasoning can be applied as above to conclude that $x_{1 \cdot k} = 0$. From this we conclude the entire sequence is zero identically.


It remains to show $$S(m) = \sum_{n \in N \setminus A(p_1, \dots, p_m)} x_n = 0.$$ Because $\sum_n x_n = 0$, this is equivalent to showing $$S(m) = \sum_{n \in A(p_1, \dots, p_m)} x_n = 0.$$

Using inclusion exclusion formula,

$$S(m) = \sum_{i=1}^m \left( (-1)^{i-1} \sum_{I \subseteq \{p_1, \dots, p_m\}, |I|=k} \sum_{n \in A(I)} x_n \right)$$

Where the innermost sum is over all numbers that are multiples of a subset of the primes in $\{p_1, \dots, p_m\}$. But this sum is of the form $\sum_{n} x_{nk}$ for $k$ the product of the primes in the subset, by assumption our sum is zero. Hence the entire expression for $S(m)$ is zero. This proves our claim, which then is used to prove our initial claim.

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