Test the convergence of the double series: sum from i,k=1
Test the convergence of the double series: $$ \sum_{i,k=1}^{\infty} \frac1{i^pk^q}. $$
Explanation of Method
Since the summand factors exactly as a product $1/i^p\cdot1/k^q$, the double sum over non-negative terms splits into the product of two independent $p$-series, each governed by the familiar $p$-series test. Convergence of the double series is therefore equivalent to convergence of both single series simultaneously.
- Factor the double series into a product of two single series.
- Apply the $p$-series test to each factor.
- Combine the two conditions and express the sum via the Riemann zeta function.
Step 1 · Factor the double series
For non-negative terms, the order of summation is immaterial, so: $$ \sum_{i,k=1}^{\infty}\frac1{i^pk^q} = \left(\sum_{i=1}^{\infty}\frac1{i^p}\right)\left(\sum_{k=1}^{\infty}\frac1{k^q}\right). $$
Step 2 · Apply the $p$-series test to each factor
$\sum1/i^p$ converges iff $p>1$; $\sum1/k^q$ converges iff $q>1$.
Step 3 · Combine
The product of two series converges iff both converge, so the double series converges iff $p>1$ and $q>1$; if either fails, at least one factor is infinite, and the double series diverges. When convergent: $\sum_{i,k=1}^{\infty}1/(i^pk^q)=\zeta(p)\zeta(q)$.
Summary Table
| Condition | Conclusion |
|---|---|
| $p>1$ and $q>1$ | Converges to $\zeta(p)\zeta(q)$ |
| $p\le1$ or $q\le1$ | Diverges |
Additional Notes
- This is the simplest possible double series, chosen deliberately to establish the baseline technique (factoring a separable summand) before Problem 148 tackles the genuinely coupled case $1/(i+k)^p$, where no such clean factorization exists.
- The special values $\zeta(2)=\pi^2/6$ and $\zeta(4)=\pi^4/90$ make $p=q=2$ a concrete numeric example: $\sum_{i,k}1/(i^2k^2)=\pi^4/36$.
- Because the summand is non-negative, Tonelli's theorem (the non-negative-term analogue of Fubini's theorem) guarantees the double sum is independent of summation order — a property that fails dramatically for the signed double series in Problem 150.
Result
The double series converges if and only if $p>1$ and $q>1$: