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: $$ \sum_{n=2}^{\infty} \frac{1}{\sqrt{n}} \ln\left(\frac{n+1}{n-1}\right). $$
Explanation of Method
Splitting the logarithm and Taylor-expanding $\ln(1\pm1/n)$ shows the linear terms of $\ln(n+1)$ and $\ln(n-1)$ reinforce rather than cancel, giving $\ln\frac{n+1}{n-1}\sim\frac2n$. Multiplying by $\frac1{\sqrt n}$ then gives an overall order of $n^{-3/2}$, comfortably inside the convergent regime.
- Expand $\ln(n+1)-\ln(n-1)$ asymptotically.
- Combine with the $1/\sqrt n$ factor to find the overall order.
- Apply the limit comparison test against $\sum 1/n^{3/2}$.
Step 1 · Expand the logarithm
Using $\ln(1+x)=x-\frac{x^2}2+O(x^3)$ with $x=\pm\frac1n$:
$$ \ln(n+1)-\ln(n-1) = \left[\ln n+\frac1n-\frac1{2n^2}\right] - \left[\ln n-\frac1n-\frac1{2n^2}\right] + O\!\left(\frac1{n^3}\right) = \frac2n+O\!\left(\frac1{n^3}\right). $$
Step 2 · Combine with the $1/\sqrt n$ factor
$$ a_n = \frac1{\sqrt n}\ln\left(\frac{n+1}{n-1}\right) = \frac1{\sqrt n}\left[\frac2n+O\!\left(\frac1{n^3}\right)\right] = \frac{2}{n^{3/2}}+O\!\left(\frac1{n^{7/2}}\right) \sim \frac{2}{n^{3/2}}. $$
Step 3 · Limit comparison test
With $b_n=\dfrac1{n^{3/2}}$: using $\lim_{n\to\infty} n\ln\left(\frac{n+1}{n-1}\right)=\lim_{h\to0^+}\frac{\ln(1+h)-\ln(1-h)}{h}=2$ (via $\lim_{x\to0}\frac{\ln(1+x)}x=1$),
$$ \lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty} n\ln\left(\frac{n+1}{n-1}\right) = 2. $$
Positive and finite; since $\sum\frac1{n^{3/2}}$ converges ($p=\frac32>1$), the series converges too.
Summary Table
| Quantity | Value |
|---|---|
| $\ln\frac{n+1}{n-1}$ asymptotics | $\sim\dfrac2n$ |
| General term $a_n$ | $\sim\dfrac{2}{n^{3/2}}$ |
| Effective exponent | $\dfrac32>1$ |
| Conclusion | Converges |
Additional Notes
- The exact limit $\lim_{n\to\infty}n\ln\frac{n+1}{n-1}=2$ can be verified two ways — via Taylor expansion or via the standard limits $\lim n\ln(1+\frac1n)=1$ and $\lim n\ln(1-\frac1n)=-1$, whose difference gives $1-(-1)=2$ — both agreeing exactly.
- This problem shows how a logarithmic factor combined with a fractional power can still land comfortably above the $p=1$ convergence threshold: the logarithm here effectively contributes an extra factor of $n^{-1}$, upgrading $n^{-1/2}$ to $n^{-3/2}$.
- Contrast with Problem 43 ($\sum1/(\ln n)^\alpha$), where logarithmic decay in the denominator is far too slow for convergence at any $\alpha$ — here the logarithm instead multiplies the term and boosts its decay rate, producing the opposite effect.
Result
The series converges: