If i have $\sum_{n=1}^\infty a_n=A$ and $\sum_{n=1}^\infty b_n=B$, can i write $\sum_{n=1}^\infty a_n\cdot b_n$ in function of $A$ and $B$ ?
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$\begingroup$Not generally. Consider the following two cases where $A = B = 1$.
Case 1: $a_1 = b_1 = 1$, and $a_n = b_n = 0$ for all $n > 1$. In this case, $\sum a_n b_n = 1$.
Case 2: $a_1 = b_2 = 1$, $a_2 = b_1 = 0$, and $a_n = b_n = 0$ for all $n > 2$. In this case, $\sum a_n b_n = 0$.
Thus, we can't come up with a formula for $\sum a_n \cdot b_n$ strictly as a function of $A$ and $B$.
$\endgroup$ $\begingroup$The simple answer is no.
But a similar trick can be used to derive Euler's reflection formula (for multivariate zeta functions)
Consider the product \begin{eqnarray*} \zeta(a) \zeta(b) = \left( \sum_{m=1}^{\infty} \frac{1}{m^a} \right) \left( \sum_{n=1}^{\infty} \frac{1}{n^b} \right) \end{eqnarray*} Now split the sum into $3$ parts $n=m,n<m$ and $n>m$ which gives \begin{eqnarray*} \zeta(a) \zeta(b) = \sum_{m=1}^{\infty} \frac{1}{m^a}\frac{1}{m^b} + \sum_{m>n>0} \frac{1}{m^a} \frac{1}{n^b} + \sum_{n>m>0} \frac{1}{m^a} \frac{1}{n^b} \\ \end{eqnarray*} and we have \begin{eqnarray*} \zeta(a) \zeta(b) = \zeta(a+b) +\zeta(a,b)+\zeta(b,a). \end{eqnarray*}
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