Prove that: sum from m,n=2 to infinity
Prove that: $$ \sum_{m,n=2}^{\infty} \frac1{(p+n)^m}=\frac1{p+1} \qquad (p>-1). $$
Explanation of Method
All terms are positive (since $p+n>0$ for $p>-1$, $n\ge2$), so the order of summation is immaterial. Summing the geometric series over $m$ first for fixed $n$ produces $1/[(p+n)(p+n-1)]$, and the resulting single series over $n$ telescopes via partial fractions.
- Sum the inner geometric series over $m$ for fixed $n$.
- Decompose the result via partial fractions.
- Evaluate the telescoping outer sum over $n$.
Step 1 · Inner geometric sum
For fixed $n\ge2$, with ratio $\frac1{p+n}$ (which has magnitude $<1$ since $p+n>p+2>1$): $$ \sum_{m=2}^{\infty}\frac1{(p+n)^m} = \frac{1/(p+n)^2}{1-1/(p+n)} = \frac1{(p+n)(p+n-1)}. $$
Step 2 · Partial fractions
$\dfrac1{(p+n)(p+n-1)}=\dfrac1{p+n-1}-\dfrac1{p+n}$.
Step 3 · Telescoping outer sum
$$ \sum_{n=2}^{\infty}\left(\frac1{p+n-1}-\frac1{p+n}\right) = \lim_{N\to\infty}\left(\frac1{p+1}-\frac1{p+N}\right) = \frac1{p+1}. $$ Verified for $p=1$: sum equals $\frac12$, matching $\frac1{p+1}=\frac12$; for $p=0$: sum equals $1=\frac1{0+1}$.
Summary Table
| Step | Result |
|---|---|
| Inner sum | $1/[(p+n)(p+n-1)]$ |
| Outer sum | Telescopes to $1/(p+1)$ |
Additional Notes
- This is the $p$-shifted generalization of Problem 149's identity $\sum_{m,n\ge2}1/m^n=1$ (the case $p=0$), extending the same technique with a translated denominator.
- Since the summand is non-negative throughout, Tonelli's theorem guarantees the double sum is well-defined regardless of order, a property essential to the "sum over $m$ first" strategy used here.
- The requirement $p>-1$ is exactly what's needed to keep $p+n>0$ for all $n\ge2$ (in fact $p+2>1$), ensuring the inner geometric series has ratio strictly less than $1$ in magnitude.
Result
The identity is proven: