Test the absolute and conditional convergence
Test the absolute and conditional convergence of the series: $$ \sum_{n=1}^{\infty}(-1)^n \frac{\ln (\sqrt{2})^n}{\sqrt{n}}. $$
Explanation of Method
The numerator simplifies drastically: $\ln(\sqrt2)^n=n\ln\sqrt2=\frac{n\ln2}2$, so the whole term is $(-1)^nk\sqrt n$ for a positive constant $k=\frac{\ln2}2$ — the terms actually grow without bound rather than shrink. Since a necessary condition for convergence of any series is that its terms tend to $0$, this series fails immediately, regardless of the alternating sign.
- Simplify $\ln(\sqrt2)^n$ to $\frac{n\ln2}2$.
- Check whether the general term tends to $0$.
- Apply the divergence (n-th term) test.
Step 1 · Simplify the general term
$$ \ln(\sqrt2)^n = n\ln(2^{1/2}) = \frac{n\ln2}2 \implies a_n = (-1)^n\frac{n\ln2}{2\sqrt n} = (-1)^n\frac{\ln2}2\sqrt n. $$
Step 2 · Check the necessary condition
With $k=\frac{\ln2}2>0$: $|a_n|=k\sqrt n\to\infty$, so $a_n$ does not tend to $0$ — in fact its magnitude is unbounded.
Step 3 · Apply the divergence test
Since $\lim_{n\to\infty}a_n\ne0$ (the terms don't even stay bounded), the series diverges by the $n$-th term test. Neither absolute nor conditional convergence is possible: $\sum|a_n|=k\sum\sqrt n$ diverges (comparable to a "$p$-series" with $p=-\frac12<1$), and Leibniz's test cannot apply either, since $b_n=k\sqrt n$ is increasing, not decreasing, and does not tend to $0$.
Summary Table
| Test | Result |
|---|---|
| Necessary condition ($a_n\to0$) | Fails: $|a_n|=k\sqrt n\to\infty$ |
| Leibniz's test | Fails: $b_n$ increasing, not $\to0$ |
| Conclusion | Diverges |
Additional Notes
- This problem is a cautionary tale about not assuming an alternating sign automatically implies some form of convergence: the sign pattern $(-1)^n$ is irrelevant once the magnitude itself fails to shrink to zero.
- The disguised growth here — $\ln(\sqrt2)^n$ looking superficially like a bounded logarithm but actually being linear in $n$ — is a deliberate trap; always simplify $\ln(\text{expression})^n$ using $\ln(x^n)=n\ln x$ before analyzing asymptotic behavior.
- Contrast with the correctly-behaved alternating series in Problems 103, 104, 107, and 108, where the underlying positive sequence genuinely decreases to $0$ — the structural difference between those and this problem is exactly what makes Leibniz's test applicable or not.
Result
The series diverges (neither absolutely nor conditionally convergent):