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: $$ \sum_{n=1}^{\infty} \frac{1}{n\sqrt{n+1}}. $$
Explanation of Method
For large $n$, $\sqrt{n+1}$ behaves essentially like $\sqrt n$, so the general term $\frac1{n\sqrt{n+1}}$ behaves like $\frac1{n^{3/2}}$ — a convergent $p$-series since $\frac32>1$. Making this precise with the limit comparison test (and cross-checking with a direct two-sided comparison) confirms that the effective decay exponent is indeed $\frac32$, so the series converges.
- Identify the asymptotic behavior of the general term as $n\to\infty$.
- Apply the limit comparison test against $\sum 1/n^{3/2}$.
- Confirm with a direct two-sided comparison bound.
Step 1 · Identify the asymptotic behavior
For large $n$, $n+1\approx n$, so $\sqrt{n+1}\approx\sqrt n$ and:
$$ a_n = \frac{1}{n\sqrt{n+1}} \approx \frac{1}{n\cdot\sqrt n} = \frac{1}{n^{3/2}}. $$
This suggests comparison with the $p$-series $\sum \frac1{n^{3/2}}$, convergent since $\frac32>1$.
Step 2 · Limit comparison test
With $b_n=\frac1{n^{3/2}}$:
$$ \lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{n^{3/2}}{n\sqrt{n+1}} = \lim_{n\to\infty}\sqrt{\frac{n}{n+1}} = \sqrt{\frac1{1+0}} = 1. $$
Since this limit is positive and finite, and $\sum\frac1{n^{3/2}}$ converges, by the limit comparison test $\sum\frac1{n\sqrt{n+1}}$ converges too.
Step 3 · Direct two-sided comparison
Since $n\le n+1\le2n$ for $n\ge1$, we get $\sqrt n\le\sqrt{n+1}\le\sqrt2\cdot\sqrt n$, and therefore:
$$ \frac{1}{\sqrt2}\cdot\frac{1}{n^{3/2}} \le \frac{1}{n\sqrt{n+1}} \le \frac{1}{n^{3/2}}. $$
The upper bound alone, via the comparison test against the convergent $\sum 1/n^{3/2}$, is enough to establish convergence directly.
Summary Table
| Quantity | Value / behavior |
|---|---|
| Effective exponent | $\frac32$ |
| Limit comparison ratio (vs. $1/n^{3/2}$) | $1$ |
| Two-sided bound | $\frac1{\sqrt2}\cdot\frac1{n^{3/2}} \le a_n \le \frac1{n^{3/2}}$ |
| Conclusion | Converges |
Additional Notes
- The essential step is recognizing that the "$+1$" inside the square root is asymptotically negligible — it shifts the value of $\sqrt{n+1}$ by a vanishing relative amount as $n\to\infty$, which is exactly why the limit comparison ratio evaluates to the clean value $1$.
- An equivalent way to see the exponent $\frac32$: writing $a_n = \frac1{\sqrt{n^2(n+1)}} = \frac1{\sqrt{n^3+n^2}}$, and $n^3+n^2\sim n^3$ for large $n$, giving $a_n\sim n^{-3/2}$ directly.
- This problem illustrates a common pattern in convergence testing: whenever the general term is a ratio or root of polynomials (or near-polynomials) in $n$, finding the overall degree/exponent and comparing to the corresponding $p$-series settles convergence quickly.
Result
The series converges: