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: $$ \sum_{n=2}^{\infty} (\sqrt{n+1}-\sqrt{n})^a \ln\frac{n-1}{n+1}. $$
Explanation of Method
Rationalizing $\sqrt{n+1}-\sqrt n$ gives order $n^{-1/2}$, so raising to the power $a$ gives order $n^{-a/2}$; the logarithm factor $\ln\frac{n-1}{n+1}$ separately behaves like $-\frac2n$, order $n^{-1}$. Multiplying these two pieces gives a combined effective exponent of $\frac{a+2}2$, and comparing to $1$ pins down exactly which values of $a$ give convergence.
- Find the asymptotic order of $(\sqrt{n+1}-\sqrt n)^a$.
- Find the asymptotic order of $\ln\frac{n-1}{n+1}$.
- Combine both orders and apply the limit comparison test to find the convergence threshold for $a$.
Step 1 · Order of $(\sqrt{n+1}-\sqrt n)^a$
Rationalizing: $\sqrt{n+1}-\sqrt n = \dfrac1{\sqrt{n+1}+\sqrt n}\sim\dfrac1{2\sqrt n}$, so $(\sqrt{n+1}-\sqrt n)^a \sim \dfrac1{2^an^{a/2}}$.
Step 2 · Order of the logarithm
Expanding $\ln(n-1)-\ln(n+1)$ via Taylor series (as in Problem 36, with opposite sign) gives $\ln\dfrac{n-1}{n+1}\sim-\dfrac2n$.
Step 3 · Combine and find the threshold
The general term satisfies $|a_n| \sim \dfrac{1}{2^{a-1}n^{(a+2)/2}}$, so the effective exponent is $\alpha=\dfrac{a+2}2$. Convergence of the comparison $p$-series requires $\alpha>1$:
$$ \frac{a+2}2 > 1 \iff a>0. $$
The limit comparison ratio against $b_n=n^{-(a+2)/2}$ works out to $\dfrac1{2^{a-1}}$, positive and finite for every $a$, confirming the threshold is exact. At the boundary $a=0$, the series reduces to the telescoping sum $\sum[\ln(n-1)-\ln(n+1)]$, whose partial sum is $\ln1+\ln2-\ln N-\ln(N+1)\to-\infty$, confirming divergence there too.
Summary Table
| Quantity | Value |
|---|---|
| $(\sqrt{n+1}-\sqrt n)^a$ | $\sim \dfrac1{2^an^{a/2}}$ |
| $\ln\frac{n-1}{n+1}$ | $\sim -\dfrac2n$ |
| Effective exponent | $\dfrac{a+2}2$ |
| Convergence | iff $a>0$ |
Additional Notes
- This problem is essentially Problem 32's rationalization technique combined with Problem 36's logarithm expansion, chained together into a single asymptotic estimate — a good illustration of how these building-block techniques compose.
- The boundary case $a=0$ is special: the power term vanishes to $1$, and the series collapses into a purely telescoping logarithmic sum, which can be evaluated exactly rather than merely estimated — and it diverges to $-\infty$, consistent with the general formula's prediction at the threshold.
- Since $\ln\frac{n-1}{n+1}<0$ for $n\ge2$ (as $n-1<n+1$), every term of the series is negative when $a>0$; the series converges to a negative value in that regime.
Result
The convergence of the series depends on $a$: