Rewriting a double summation. If I am given this function: $$f(x) = \sum_{i = 1}^{\infty} \frac{1}{i^x}$$
Is there a way to rewrite:
$$g(x) = \sum_{ j = 1}^{\infty} \sum_{i = j}^{\infty} \frac{1}{(i \cdot j)^x}$$
In terms of f(x). By simple arthimatic I know that $f(x)^2 = 2 \cdot g(x) - f(2x)$.
The question is if for a more general way to rewrite the summation. If $g(x)$ contained 3 summation with $i,j,k$ how would you rewrite it in terms of $f(x)$.
 A: Let
\begin{eqnarray*}
f_1(x) &=&  \sum_{i \geq 1} \frac{1}{i^x} \\ 
f_2(x) &=& \sum_{i >j \geq 1} \frac{1}{(ij) ^x} \\ 
f_3(x) &=& \sum_{i>j>k \geq 1} \frac{1}{(ijk)^x}. \\ 
\end{eqnarray*}
In your question you have calculated
\begin{eqnarray*}
\left( 1 + \frac{1}{2^x} + \frac{1}{3^x} + \cdots \right) ^2 &=& \sum_{i \geq 1} \frac{1}{i^{2x} } + 2 \sum_{i >j \geq 1} \frac{1}{(ij) ^x} \\
(f_1(x))^2 &=& f_1(2x) +2 f_2(x).
\end{eqnarray*}
Calculating similarly 
\begin{eqnarray*}
\left( 1 + \frac{1}{2^x} + \frac{1}{3^x} + \cdots \right) ^3 = \sum_{i \geq 1} \frac{1}{i^{3x} } + 3 \sum_{i >j \geq 1} \frac{1}{(i^2j) ^x} + 3 \sum_{i >j \geq 1} \frac{1}{(ij^2) ^x} + 6 \sum_{i >j>k \geq 1} \frac{1}{(ijk) ^x} \\
\left( 1 + \frac{1}{2^x} + \frac{1}{3^x} + \cdots \right) \left( 1 + \frac{1}{2^{2x}} + \frac{1}{3^{2x}} + \cdots \right) = \sum_{i \geq 1} \frac{1}{i^{3x} } +  \sum_{i >j \geq 1} \frac{1}{(i^2j) ^x} +  \sum_{i >j \geq 1} \frac{1}{(ij^2) ^x} .  \\
\end{eqnarray*}
Multiply the second equation by $3$ and subtract it from the first gives
\begin{eqnarray*}
(f_1(x))^3  -3f_1(x) f_1(2x) = -2f_1(3x) +6 f_3(x).
\end{eqnarray*}
A: For
$g(x) 
= \sum_{ j = i}^{\infty} \sum_{i = 0}^{\infty} \dfrac{1}{(i \cdot j)^x}
$
and sums like it,
anything independent
of the innermost index of summation
can be pulled out.
However,
you have a mistake in
the way you have written this:
An outer summation
can not depend
on more inner indices.
This should be written
$g(x) 
= \sum_{i = 1}^{\infty}\sum_{ j = i}^{\infty}  \dfrac{1}{(i \cdot j)^x}
$.
Also note that
$i$ must start at $1$
since
$\dfrac1{i^x}$
is not defined at $i = 0$.
In this case,
you can do
$g(x) 
=  \sum_{i = 1}^{\infty}\sum_{ j = i}^{\infty} \dfrac{1}{(i \cdot j)^x}
=  \sum_{i = 1}^{\infty}\dfrac1{i^x}\sum_{ j = i}^{\infty} \dfrac{1}{j^x}
$.
You can also reverse the order of summation
to get
$g(x) 
=  \sum_{i = 1}^{\infty}\sum_{ j = i}^{\infty} \dfrac{1}{(i \cdot j)^x}
=  \sum_{ j = 1}^{\infty}\sum_{i = 1}^{j} \dfrac{1}{(i \cdot j)^x}
=  \sum_{ j = 1}^{\infty}\dfrac1{j^x}\sum_{i = 1}^{j} \dfrac{1}{i^x}
$.
