Examine the convergence of the series: sum from n=1
Examine the convergence of the series: $$ \sum_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}+\ln \frac{n+1}{n}\right). $$
Explanation of Method
The logarithmic part telescopes exactly: $\sum_{n=1}^N\left[\ln(n+1)-\ln n\right]=\ln(N+1)$. Adding this to the divergent partial sum of $\sum1/\sqrt n$ shows the total partial sum grows like $2\sqrt N+\ln N\to\infty$. Equivalently, since both pieces of the general term are asymptotically positive and the dominant piece $1/\sqrt n$ alone already gives a divergent $p$-series, the whole series diverges by the limit comparison test.
- Telescope the logarithmic part of the partial sum.
- Combine with the divergent $p$-series part to show $S_N\to\infty$.
- Confirm via the limit comparison test against $1/\sqrt n$.
Step 1 · Telescope the logarithmic part
$$ \sum_{n=1}^{N} \ln\frac{n+1}n = \sum_{n=1}^N\left[\ln(n+1)-\ln n\right] = \ln(N+1) - \ln1 = \ln(N+1). $$
Step 2 · Combine with the $p$-series part
The $N$-th partial sum is:
$$ S_N = \sum_{n=1}^N\frac1{\sqrt n} + \ln(N+1). $$
Since $\sum\frac1{\sqrt n}$ is a divergent $p$-series ($p=\frac12<1$), $\sum_{n=1}^N\frac1{\sqrt n}\sim2\sqrt N\to\infty$, and $\ln(N+1)\to\infty$ too, so $S_N\to\infty$.
Step 3 · Confirm via limit comparison
With $a_n=\frac1{\sqrt n}+\ln\frac{n+1}n$ and $v_n=\frac1{\sqrt n}$, using $\ln\left(1+\frac1n\right)\sim\frac1n$ for large $n$:
$$ \frac{a_n}{v_n} = 1+\sqrt n\ln\left(1+\frac1n\right) \sim 1+\frac1{\sqrt n} \longrightarrow 1. $$
Since the limit is a finite nonzero constant and $\sum\frac1{\sqrt n}$ diverges, so does $\sum a_n$.
Summary Table
| Component | Behavior |
|---|---|
| $\sum\frac1{\sqrt n}$ (partial sum) | $\sim2\sqrt N\to\infty$ |
| $\sum\ln\frac{n+1}n$ (telescoping) | $=\ln(N+1)\to\infty$ |
| $\lim a_n/v_n$ | $1$ |
| Conclusion | Diverges |
Additional Notes
- Since both components of the general term are individually positive and each already forces divergence on its own, there is no possibility of cancellation here — unlike series formed as a difference of two divergent parts, where delicate cancellation can sometimes restore convergence.
- The telescoping identity $\sum_{n=1}^N\ln\frac{n+1}n=\ln(N+1)$ is the same mechanism used throughout the logarithmic-telescoping problems early in this collection; here it plays a secondary role, since the dominant $\frac1{\sqrt n}$ term alone already determines the outcome.
- This problem is a useful reminder that adding a convergent-looking correction (the logarithmic term, which individually forms a divergent telescoping series growing only logarithmically) to an already-divergent $p$-series term cannot rescue convergence — the faster-growing term always dominates the overall behavior.
Result
The series diverges: