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: $$ \sum_{n=1}^{\infty} \left(\frac{1}{n} - \ln\frac{n+1}{n}\right). $$
Explanation of Method
Writing $\ln\frac{n+1}n=\ln(1+\frac1n)$ and Taylor-expanding, the leading term $\frac1n$ cancels exactly against the subtracted $\frac1n$, leaving a remainder of order $\frac1{2n^2}$. That quadratic decay places the series well within convergent territory, confirmed by both a limit comparison test and an independent L'Hôpital computation.
- Expand $\ln(1+\frac1n)$ via Taylor series and cancel the leading term.
- Apply the limit comparison test against $\sum1/n^2$.
- Verify the limit independently using L'Hôpital's Rule.
Step 1 · Expand and cancel the leading term
$\ln\frac{n+1}n=\ln(1+\frac1n)=\frac1n-\frac1{2n^2}+\frac1{3n^3}-\cdots$, so:
$$ a_n = \frac1n-\ln\!\left(1+\frac1n\right) = \frac1{2n^2}-\frac1{3n^3}+\cdots \sim \frac1{2n^2}. $$
Step 2 · Limit comparison test
With $b_n=\dfrac1{n^2}$:
$$ \lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{\frac1{2n^2}+O(1/n^3)}{1/n^2} = \frac12. $$
Positive and finite; since $\sum\frac1{n^2}$ converges, the series converges too.
Step 3 · Verify via L'Hôpital's Rule
Setting $h=\frac1n\to0^+$, the same limit is $\lim_{h\to0^+}\frac{h-\ln(1+h)}{h^2}$, of form $\frac00$. Applying L'Hôpital's Rule twice:
$$ \lim_{h\to0^+}\frac{1-\frac1{1+h}}{2h} = \lim_{h\to0^+}\frac{h/(1+h)}{2h} = \lim_{h\to0^+}\frac{1}{2(1+h)} = \frac12, $$
confirming the earlier result.
Summary Table
| Quantity | Value |
|---|---|
| General term $a_n$ | $\sim\dfrac1{2n^2}$ |
| Limit comparison ratio (vs. $1/n^2$) | $\dfrac12$ |
| Conclusion | Converges |
Additional Notes
- This series is closely tied to Euler's constant $\gamma$: the sum $\sum_{n=1}^{N}\left(\frac1n-\ln\frac{n+1}n\right)$ telescopes the logarithmic part into $\ln(N+1)$, and the resulting identity $H_N-\ln(N+1)\to\gamma$ is one standard way of defining the Euler–Mascheroni constant.
- The cancellation of the $\frac1n$ term mirrors what happened in Problem 35 (with exponentials instead of logarithms): whenever a natural approximation and the exact quantity share the same leading-order term, subtracting them boosts the decay rate of the remainder by (at least) one power of $n$.
- All terms here are positive (since $\ln(1+x)<x$ for $x>0$), so the partial sums increase monotonically toward the finite limit, consistent with convergence to a value related to $\gamma$.
Result
The series converges: