Solutions - 0180-03

The series has positive terms.

Notice that $\sqrt{n}$ is much greater than $\ln n$ for large $n$. So we anticipate that this term will dominate, and we compare the series to ${\displaystyle \frac{1}{\sqrt{n}}}$.

$$\begin{array}{c}{a}_n\cdot b_n^{-1}\gg \gg \frac{1}{\sqrt{n}+\ln n}\cdot \sqrt{n}\\ \\ \gg \gg \frac{n^{-1/2}}{n^{-1/2}}\cdot \frac{n^{1/2}}{n^{1/2}+\ln n}\gg \gg \frac{1}{1+{\displaystyle \frac{\ln n}{n^{1/2}}}}\end{array}$$

We seek the limit of this as $n\to \infty$. Apply L’Hopital’s Rule to the fraction:

$$\frac{\ln x}{x^{1/2}}{\underset{d/dx}{⟶}}^{d/dx}\frac{1/x}{\frac{1}{2\sqrt{x}}}\gg \gg \frac{2}{\sqrt{x}}\stackrel{x\to \infty }{⟶}0$$

Therefore, by continuity:

$$\frac{1}{1+{\displaystyle \frac{\ln n}{n^{1/2}}}}\stackrel{n\to \infty }{⟶}\frac{1}{1+0}\gg \gg 1=L$$

Since $0<L<\infty$, the LCT says that both series converge or diverge.

Since ${\displaystyle \sum \frac{1}{\sqrt{n}}}$ diverges ($p=1/2$), the original series must diverge.