Prove that the double series sum from i,k=1 to infinity
Prove that the double series $\displaystyle\sum_{i,k=1}^{\infty} \frac1{(i+k)^p}$ converges for $p>2$.
Explanation of Method
Unlike Problem 147's separable summand, $1/(i+k)^p$ couples $i$ and $k$, so the double sum must be reorganized by the value $n=i+k$: for fixed $n$, there are exactly $n-1$ pairs $(i,k)$ with $i,k\ge1$, $i+k=n$. This turns the double series into a single series $\sum(n-1)/n^p$, resolvable by the ordinary $p$-series test, giving the threshold $p>2$. An independent AM-GM comparison confirms the same threshold.
- Reindex by $n=i+k$, counting the $n-1$ pairs for each fixed $n$.
- Split $\sum(n-1)/n^p$ into two $p$-series and apply the $p$-series test.
- Confirm via an independent AM-GM comparison argument.
Step 1 · Reindex by $n=i+k$
For fixed $n\ge2$, the pairs $(i,k)$ with $i,k\ge1$ and $i+k=n$ number exactly $n-1$ (namely $i=1,\ldots,n-1$). So: $$ \sum_{i,k=1}^{\infty}\frac1{(i+k)^p} = \sum_{n=2}^{\infty}\frac{n-1}{n^p}. $$
Step 2 · Split and apply the $p$-series test
$$ \sum_{n=2}^{\infty}\frac{n-1}{n^p} = \sum_{n=2}^{\infty}\frac1{n^{p-1}} - \sum_{n=2}^{\infty}\frac1{n^p}. $$ The first converges iff $p-1>1$, i.e. $p>2$; the second converges iff $p>1$. The more restrictive condition, $p>2$, governs the whole difference; for $1<p\le2$ or $p\le1$, at least the first (or both) series diverges, so the double series diverges.
Step 3 · Confirm via AM-GM comparison
Since $i+k\ge2\sqrt{ik}$: $\dfrac1{(i+k)^p}\le\dfrac1{2^p(ik)^{p/2}}$, so $$ \sum_{i,k=1}^{\infty}\frac1{(i+k)^p} \le \frac1{2^p}\left(\sum_{i=1}^{\infty}\frac1{i^{p/2}}\right)^2, $$ and the right side converges iff $p/2>1$, i.e. $p>2$ — matching Step 2's threshold exactly.
Summary Table
| Range of $p$ | Conclusion |
|---|---|
| $p>2$ | Converges |
| $p\le2$ | Diverges |
Additional Notes
- The threshold $p>2$ (rather than $p>1$ as for the separable series of Problem 147) reflects the "coupling" of $i$ and $k$: since roughly $n-1\sim n$ pairs contribute at each level $n=i+k$, an extra power of $n$ is effectively absorbed compared to a single-index $p$-series, raising the convergence threshold by exactly $1$.
- The AM-GM comparison in Step 3 is a standard technique for bounding sums over "coupled" indices by a product of separable sums, and reappears whenever a double series' summand depends on $i+k$, $ik$, or similar symmetric combinations rather than $i$ and $k$ independently.
- This problem, together with Problem 147, establishes the two basic patterns for two-index series (separable vs. sum-coupled); Problem 150's double series with the $(i-j)$ factor introduces a third, genuinely asymmetric pattern where even the order of summation affects the answer.
Result
The double series converges precisely for $p>2$: