How to express a sum of two series in terms of known sums? 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 $B$ ?
 A: 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 $B$.  
A: The simple answer is no. 
But a similar trick can be used to derive Euler's reflection formula (for multivariate zeta functions) https://en.wikipedia.org/wiki/Multiple_zeta_function#Euler_reflection_formula
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*}
