Rewriting a double summation.

Question

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 $f(x)$.

Answer 1

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*}

Answer 2

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} $.