Knowing that sum from n=1 to infinity 1/(n^()alpha)
Knowing that $\sum_{n=1}^{\infty}\frac{1}{n^\alpha}$ converges for $\alpha>1$ and diverges for $\alpha\le1$, determine the convergence of: $$ \sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt[3]{n}}. $$
Explanation of Method
Rationalizing $\sqrt{n+1}-\sqrt n$ turns it into $\frac1{\sqrt{n+1}+\sqrt n}\sim\frac1{2\sqrt n}$, a term of order $n^{-1/2}$. Dividing by $\sqrt[3]n=n^{1/3}$ combines the exponents to give an overall order of $n^{-5/6}$ — below the convergence threshold of $1$ — so the limit comparison test with $\sum n^{-5/6}$ shows divergence.
- Rationalize the numerator to reveal its asymptotic order.
- Combine exponents to find the effective decay rate.
- Apply the limit comparison test against $\sum 1/n^{5/6}$.
Step 1 · Rationalize the numerator
$$ \sqrt{n+1}-\sqrt n = \frac{(n+1)-n}{\sqrt{n+1}+\sqrt n} = \frac{1}{\sqrt{n+1}+\sqrt n}. $$
So the general term is $a_n = \dfrac{1}{(\sqrt{n+1}+\sqrt n)\cdot n^{1/3}}$.
Step 2 · Find the effective exponent
For large $n$: $\sqrt{n+1}+\sqrt n\approx2\sqrt n=2n^{1/2}$, so:
$$ a_n \approx \frac{1}{2n^{1/2}\cdot n^{1/3}} = \frac{1}{2n^{5/6}}, \qquad \frac12+\frac13=\frac56<1. $$
Step 3 · Limit comparison test
With $b_n=\dfrac1{n^{5/6}}$:
$$ \lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{n^{1/2}}{\sqrt{n+1}+\sqrt n} = \lim_{n\to\infty}\frac{1}{\sqrt{1+\frac1n}+1} = \frac12. $$
Positive and finite; since $\sum\frac1{n^{5/6}}$ diverges ($p=\frac56<1$), by the limit comparison test the original series diverges too.
Summary Table
| Quantity | Value |
|---|---|
| Rationalized numerator | $\sqrt{n+1}-\sqrt n = \dfrac1{\sqrt{n+1}+\sqrt n}$ |
| Effective exponent | $\dfrac12+\dfrac13=\dfrac56<1$ |
| Limit comparison ratio | $\dfrac12$ |
| Conclusion | Diverges |
Additional Notes
- Rationalizing $\sqrt{n+1}-\sqrt n$ is essential here: left unrationalized, the numerator's decay rate is not visible, since both $\sqrt{n+1}$ and $\sqrt n$ individually grow without bound.
- Adding exponents from separate factors ($n^{1/2}$ from the rationalized numerator's denominator, $n^{1/3}$ from the given denominator) is a quick way to find the overall order of a compound expression, as long as each factor's individual asymptotic behavior is already known.
- Since $\frac56$ is close to but below $1$, this series diverges only marginally more slowly than the harmonic series — its partial sums grow, but at a rate close to $n^{1/6}$.
Result
The series diverges: