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: $$ \sum_{n=1}^{\infty} \frac{1}{\sqrt{3n(2n-1)}}. $$
Explanation of Method
Inside the square root, $3n(2n-1)$ behaves like $6n^2$ for large $n$, so the whole term behaves like $\frac{1}{\sqrt6\,n}$ — an effective exponent of exactly $1$, the harmonic-series boundary. Making that asymptotic precise with the limit comparison test (and cross-checking with a direct bound) confirms the series diverges.
- Identify the asymptotic behavior of the general term.
- Apply the limit comparison test against $\sum 1/n$.
- Confirm with a direct comparison bound.
Step 1 · Identify the asymptotic behavior
Since $2n-1\approx2n$ for large $n$: $3n(2n-1)\approx6n^2$, so $\sqrt{3n(2n-1)}\approx\sqrt6\,n$, and:
$$ a_n = \frac{1}{\sqrt{3n(2n-1)}} \approx \frac{1}{\sqrt6}\cdot\frac1n = \frac{1}{n\sqrt{6-\frac3n}}. $$
Step 2 · Limit comparison test
With $b_n=\frac1n$:
$$ \lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{n}{\sqrt{6n^2-3n}} = \lim_{n\to\infty}\frac{1}{\sqrt{6-\frac3n}} = \frac{1}{\sqrt6}. $$
Positive and finite; since $\sum\frac1n$ diverges, by the limit comparison test $\sum\frac1{\sqrt{3n(2n-1)}}$ diverges too.
Step 3 · Direct comparison confirmation
Since $2n-1<2n$: $3n(2n-1)<6n^2$, so $\sqrt{3n(2n-1)}<\sqrt6\,n$, giving:
$$ \frac{1}{\sqrt{3n(2n-1)}} > \frac{1}{\sqrt6\,n}. $$
Since $\sum\frac1{\sqrt6\,n}$ diverges, so does the original series, by direct comparison.
Summary Table
| Quantity | Value / behavior |
|---|---|
| Effective exponent | $1$ (harmonic boundary) |
| Limit comparison ratio (vs. $1/n$) | $\dfrac1{\sqrt6}$ |
| Conclusion | Diverges |
Additional Notes
- Although the denominator involves a product of two linear factors under a square root, the net effect is still just a single power of $n$, since $\sqrt{n\cdot n}=n$ — a useful shortcut for quickly estimating effective exponents in products like this.
- The constant $\frac1{\sqrt6}$ has no bearing on convergence, only on how the terms compare in magnitude to $\frac1n$; only the exponent (here, $1$) determines convergence or divergence.
- This problem is structurally similar to Problem 27's odd-harmonic series: both reduce, after simplification, to a constant multiple of the harmonic series, and both therefore diverge.
Result
The series diverges: