Help needed in understanding how a certain result was found.

Question

$$\sum_{j=1}^{a-1}\widetilde{\zeta}(j+1)\widetilde{\zeta}(2a-j)=\sum_{j=1}^{a-1}\widetilde{\zeta}(2j)\widetilde{\zeta}(2a+1-2j)$$

I'm trying to understand how the two are equal. At first, I thought they separated the sum according to the parity of $j$ but I soon discarded that thought. This step is crucial as the paper is on an identity I need to evaluate a certain integral. Any help however small will be appreciated.

Answer 1

In general, for any sequence $f(n)$ and for any $a \geq 1$, we have

$$ \sum_{j=1}^{a-1} f(j+1)f(2a-j) = \sum_{j=1}^{a-1} f(2j)f(2a+1-2j). \tag{1} $$

Indeed, write $g(j) = f(j)f(2a+1-j)$ and note that $g(j) = g(2a+1-j)$. Then substituting $j' = 2a-j-1$,

\begin{align*} 2\cdot\text{[LHS of (1)]} &= \sum_{j=1}^{a-1} g(j+1) + \sum_{j'=a}^{2a-2} g(2a-j') \\ &= \sum_{j=1}^{a-1} g(j+1) + \sum_{j'=a}^{2a-2} g(j'+1) \\ &= g(2) + g(3) + \cdots + g(2a-2) + g(2a-1) \end{align*}

On the other hand, substituting $j' = a - j$,

\begin{align*} 2\cdot\text{[RHS of (1)]} &= \sum_{j=1}^{a-1} g(2j) + \sum_{j'=1}^{a-1} g(2a - 2j') \\ &= \sum_{j=1}^{a-1} [g(2j) + g(2j+1)] \\ &= g(2) + g(3) + \cdots + g(2a-2) + g(2a-1) \end{align*}

Therefore $\text{(1)}$ holds.

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Visual Proof. For each set $I \subseteq \{1,2,3, \ldots, 2a\}$ of indices, I will visualize the sum $\sum_{i \in I} g(i)$ by a row of $2a$ circles, in which the $i$-th circle is filled if $i \in I$ (so that $g(i)$ is present in the sum) and empty if $i \notin I$ (so that $g(i)$ is absent from the sum).

Example. If $a = 4$ and $I = \{2, 3, 6\}$, then $$\sum_{i \in \{2,3,6\}} g(i) = \text{○●●○○●○○}. $$ Since $g(i) = g(9-i)$, it is also true that $$\sum_{i \in \{2,3,6\}} g(i) = \sum_{i \in \{3,6,7\}} g(i) = \text{○○●○○●●○}. $$ Note that $\text{○○●○○●●○}$ is the reverse of $\text{○●●○○●○○}$.

In general, if $s$ is the string of ○'s and ●'s representing a sum, then the reverse of $s$ also represents the same sum. Keeping this in mind, we find that

\begin{align*} \sum_{j=1}^{a-1} g(i+1) &= \text{○}\underbrace{\color{blue}{\text{●●}\cdots\text{●●}}}_{2\text{ to } a}\text{○○}\cdots\text{○○} \\ &= \text{○○}\cdots\text{○○}\underbrace{\color{red}{\text{●●}\cdots\text{●●}}}_{a+1\text{ to } 2a-1}\text{○} \tag{reversal} \\ &= \frac{1}{2}[\text{○}\color{blue}{\text{●●}\cdots\text{●●}}\color{red}{\text{●●}\cdots\text{●●}}\text{○}] \tag{averaging} \\ &= \frac{1}{2} \sum_{i=2}^{2a-1} g(i) \end{align*}

and

\begin{align*} \sum_{j=1}^{a-1} g(2j) &= \text{○}\color{blue}{\text{●}\text{○}\text{●}\cdots\text{○}\text{●}\text{○}}\text{○} \\ &= \text{○}\color{red}{\text{○}\text{●}\text{○}\text{●}\cdots\text{○}\text{●}}\text{○} \tag{reversal} \\ \\ &= \frac{1}{2}[ \text{○}\color{blue}{\text{●}}\color{red}{\text{●}}\color{blue}{\text{●}}\color{red}{\text{●}} \cdots \color{blue}{\text{●}}\color{red}{\text{●}}\text{○} ] \tag{averaging} \\ &= \frac{1}{2} \sum_{i=2}^{2a-1} g(i) \end{align*}