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Let $(x_n)$ be a real sequence such that $\sum^{\infty}_{n=1}|x_n|$ converges and, for each $k\in \mathbb N, \sum^{\infty}_{n=1}x_{kn}=0.$ Show that $x_n=0$ for all $n$.

I can't get anywhere useful.

Preferably I would ask that the answer is given from 'first principles' if that's the right term (i.e. no theorems or known results on such series) as this is supposed to be done just as the course introduces series and convergence.

Any help is appreciated

Thanks

Aka_aka_aka_ak
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  • A possible line of attack: Show that for sequences satisfying the conditions, then $x_1=0$. Then realize that the same conditions are satisfied for the sequence $x_{kn}$, and hence $x_{1\cdot n} = 0$ as well. – abnry Aug 31 '16 at 17:29
  • you will need the Möbius function, and the bound $|\sum_{n \ge k/d} x_{nd}| < \sum_{n \ge k} |x_n|$ – reuns Aug 31 '16 at 17:51
  • @user1952009 Yeah, that seems to be the way to go. On the original questions (to which this is a duplicate) I had more or less gotten to where the third answer was but had difficulty justifying it when applied to an infinite sum. The fact that it requires the Mobius function surprises me, given that this was taken from a problem sheet set to first term, first year undergraduates where the Mobius function had not been mentioned in the notes at all. – Aka_aka_aka_ak Aug 31 '16 at 18:02
  • You can see my answer to the old question that doesn't use the mobius function. It uses the inclusion-exclusion formula though. – abnry Sep 01 '16 at 23:17
  • @reuns What does you bound mean? Could you write down a partial sum version of that? After all we do not know a priori if the LHS series converges. Thank you. – Ma Joad Dec 15 '20 at 19:06

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