The problem is given below: For two sequences of complex numbers $\{a_0, a_1, \cdots, a_n, \cdots\}$ and $\{b_0, b_1, \cdots, b_n, \cdots\}$ show that the following relations are equivalent: $$a_n = \sum_{k = 0}^n{b_k}\ for\ all\ n \Leftrightarrow b_n = \sum_{k = 0}^n{(-1)^{k + n} a_k}\ for\ all\ n.$$

I was trying to learn Mobius Inversion Formula and Multiplicative functions and I found this problem. I understand how to use the formula to deal with something like $\sum_{d|n} f(d)$, but I didn't get how to solve things like $\sum_{d=0}^n f(d)$. What's the strategy when we face situations like this?

Thanks in advance.

  • $\begingroup$ Just an idea: Don't you think that induction could do the trick? $\endgroup$ – sranthrop Mar 5 '16 at 13:17
  • $\begingroup$ @sranthrop How do I perform the induction then? $\endgroup$ – SCaffrey Mar 5 '16 at 13:36
  • $\begingroup$ Are you sure that your formulas are correct? If $b_k:=k$, then $a_0=0,a_1=1,a_2=3$, but for $c_n:=\sum_{k=0}^n(-1)^{k+n}a_k$ we have $c_0=0,c_1=1,c_2=2$, so $c_n\neq b_n$. $\endgroup$ – sranthrop Mar 6 '16 at 15:54
  • $\begingroup$ $b_0=0,b_1=1,b_2=2$, and $c_0=0,c_1=1,c_2=2$ ??? Did I get it right? $\endgroup$ – SCaffrey Mar 7 '16 at 14:36
  • $\begingroup$ I'm sorry, you need to go to $n=3$: $a_3=6$, $b_3=3$, but $c_3=4$. $\endgroup$ – sranthrop Mar 7 '16 at 16:19

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