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For two sequences of complex numbers $\{a_0, a_1, ..., a_n, ...\}$ and $\{b_0, b_1, ..., b_n, ...\}.$ Show that the following relations are equivalent: \begin{equation} a_n = \sum_{k=0}^n b_k \iff b_n = \sum_{k=0}^n(-1)^{k+n}a_k \text{ for all } n. \end{equation}

I know it had to do with some type of Mobius inversion. I tried as follows

\begin{align} \sum_{k=0}^n b_k &= \sum_{k=0}^n \sum_{m=0}^k(-1)^{m+k}a_m \\ &= \sum_{m=0}^n \sum_{k=m}^n(-1)^{m+k}a_m \\ &= \sum_{m=0}^n(-1)^ma_m \sum_{k=m}^n(-1)^{k} \end{align}

I have no idea what to do next. I am not even sure that I interchange sum in correct way. Any help will be really appreciated. Thanks.

Note--I saw similar questions but they are not of so much help.

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    $\begingroup$ There seems to be a mistake? Let $a_n=n+1$ and $b_n=1$. Then $a_n=\sum_{k=0}^n b_k$, but $1\neq \sum_{k=0}^n (-1)^{k+n}(k+1)$. $\endgroup$ – Mike Earnest Feb 22 at 19:30
  • $\begingroup$ @MikeEarnest Well this is a problem form Berkeley. The problem 10 on Page 7 of math.berkeley.edu/~stankova/MathCircle/Multiplicative.pdf $\endgroup$ – Learner Feb 22 at 19:42
  • $\begingroup$ Your source is wrong, then. Obviously, $a_n=\sum_{k=0}^nb_k$ if and only if $b_n=a_n-a_{n-1}$ (except $b_0=a_0$), and in general $a_n-a_{n-1}\neq a_n-a_{n-1}+a_{n-2}-a_{n-3}+\dots+(-1)^n a_0$. $\endgroup$ – Mike Earnest Feb 22 at 19:51
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I agree the question is not correct as stated. However, I believe it's a typo on the starting value of the first sum, with it meant to be for $k = n - 1$, as this will then make the question correct. Thus, the statement to prove would instead be

$$a_n = \sum_{k=n-1}^n b_k \iff b_n = \sum_{k=0}^n(-1)^{k+n}a_k \text{ for all } n \tag{1}\label{eq1}$$

This can be shown using induction. First, note that $n = 0$, where the sum starts at $0$, we get $a_0 = b_0$ on the left & right sides. For $n = 1$, the left side gives $a_1 = b_0 + b_1$, from which $b_1 = -b_0 + a_1 = -a_0 + a_1$, which is the right side.

Next, assume this holds true for all $n \le m$ for some integer $m \ge 0$. For $n = m + 1$, the left side of \eqref{eq1} becomes

$$a_{m+1} = b_{m} + b_{m+1} \Rightarrow b_{m+1} = -b_{m} + a_{m+1} \tag{2}\label{eq2}$$

From the induction hypothesis, we get that

\begin{align} -b_{m} & = -\sum_{k=0}^m (-1)^{k+m}a_k \\ & = \sum_{k=0}^m (-1)^{k+\left(m + 1\right)}a_k \\ & = \sum_{k=0}^{m+1} (-1)^{k+\left(m + 1\right)}a_k - \left(-1\right)^{\left(m+1\right)+\left(m+1\right)}a_{m+1} \\ & = \sum_{k=0}^{m+1} (-1)^{k+\left(m + 1\right)}a_k - a_{m+1} \end{align}

Substituting this into \eqref{eq2} gives that

$$b_{m+1} = \sum_{k=0}^{m+1} (-1)^{k+\left(m + 1\right)}a_k \tag{3}\label{eq3}$$

This proves the statement one way. As all of the steps are reversible, this means the statement is also true the other way. Thus, it's also true for $n = m + 1$, completing the induction procedure.

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