Examine the convergence of the series: sum from n=1
Examine the convergence of the series: $$ \sum_{n=1}^{\infty} \frac{\sqrt{n+a}-\sqrt{n+b}}{n^\alpha}, \qquad a>0,\; b>0. $$
Explanation of Method
Rationalizing the numerator by multiplying by the conjugate $\sqrt{n+a}+\sqrt{n+b}$ converts the difference of square roots into the constant $a-b$ over a sum of square roots — a much more tractable form. Since that sum grows like $2\sqrt n$, the whole term behaves asymptotically like $\frac{a-b}{2n^{\alpha+1/2}}$, reducing the problem to a $p$-series comparison with the shifted exponent $\alpha+\frac12$.
- Rationalize the numerator using the conjugate.
- Find the asymptotic behavior of the denominator $\sqrt{n+a}+\sqrt{n+b}\sim2\sqrt n$.
- Apply the limit comparison test against $\sum n^{-(\alpha+1/2)}$.
Step 1 · Rationalize the numerator
$$ \sqrt{n+a}-\sqrt{n+b} = \frac{(n+a)-(n+b)}{\sqrt{n+a}+\sqrt{n+b}} = \frac{a-b}{\sqrt{n+a}+\sqrt{n+b}}. $$
So the general term is $u_n=\dfrac{a-b}{n^\alpha(\sqrt{n+a}+\sqrt{n+b})}$.
Step 2 · Asymptotics of the denominator
Using $\sqrt{n+c}=\sqrt n\left(1+\frac c{2n}+O(1/n^2)\right)$ for $c=a,b$:
$$ \sqrt{n+a}+\sqrt{n+b} = 2\sqrt n + O(1/\sqrt n) \sim 2\sqrt n. $$
Step 3 · Compare with a $p$-series
So $u_n\sim\dfrac{a-b}{2n^{\alpha+1/2}}$. With $v_n=n^{-(\alpha+1/2)}$:
$$ \lim_{n\to\infty}\frac{u_n}{v_n} = \lim_{n\to\infty}\frac{(a-b)\sqrt n}{\sqrt{n+a}+\sqrt{n+b}} = \frac{a-b}2, $$
a finite nonzero constant when $a\ne b$. By the limit comparison test, $\sum u_n$ converges iff $\sum n^{-(\alpha+1/2)}$ converges, i.e. iff $\alpha+\frac12>1$, i.e. $\alpha>\frac12$. (If $a=b$, every term is $0$ and the series trivially converges.)
Summary Table
| Condition | Conclusion |
|---|---|
| $a=b$ | Trivially converges (all terms $0$) |
| $a\ne b,\;\alpha>\dfrac12$ | Converges |
| $a\ne b,\;\alpha\le\dfrac12$ | Diverges |
Additional Notes
- The rationalization trick used here is the same "difference of square roots via conjugate" technique that also opened several telescoping-series problems early in this collection (Problems 2–5) — it converts an awkward difference into a well-behaved ratio ideal for asymptotic comparison.
- The precise values of $a$ and $b$ never enter the convergence threshold, only their difference (through the nonzero constant $\frac{a-b}2$) — the threshold $\alpha>\frac12$ is universal across all choices of distinct $a,b>0$.
- The half-integer shift in the threshold (from $\alpha>1$ for a plain $p$-series to $\alpha>\frac12$ here) directly reflects that the rationalized numerator contributes an extra factor of $n^{-1/2}$ worth of decay beyond the bare $n^{-\alpha}$, exactly as expected from the square-root asymptotics.
Result
For $a\ne b$, the series converges if and only if $\alpha>\frac12$: