Ratio test for convergent series - Does that mean that the series $1 +frac{1}{2} + frac{1}{3} +dotsb$ converges?

284. An infinite series is convergent if from and after some fixed term, the ratio of each term to the preceding term is numerically less than some quantity which is itself numerically less than unity.
     Let the series beginning from the fixed term be denoted by
     $$u_1+u_2+u_3+u_4+\dotsb,$$
     and let
     $$\frac{u_2}{u_1} where $r<1$.
     Then
     \begin{align*}
     
     &u_1+u_2+u_3+u_4+\dotsb\\
     &=u_1\left(1+\frac{u_2}{u_1}+\frac{u_3}{u_2}\cdot\frac{u_2}{u_1}+\frac{u_4}{u_3}\cdot\frac{u_3}{u_2}\cdot\frac{u_2}{u_1}+\dotsb\right)\\
     &\end{align*}
     that is, $<\frac{u_1}{1-r}$, since $r<1$.
     Hence the given series is convergent.

Does that mean that the series $1 +\frac{1}{2} + \frac{1}{3} +\dotsb$ should be a convergent one?
$$S_n = \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dotsb$$

Here, we have
\begin{gather}
\frac{u_n}{u_{n-1}} = \frac{1/n}{1/(n-1)} = \frac{n-1}{n} = 1-\frac{1}{n}\\
∴\boxed{\frac{u}{u_{n-1}}<1}\\
∴\text{Series should be convergent}
\end{gather}

Answer

You need $\frac{u_n}{u_{n-1}} < r < 1$ for some $r < 1$ and all $n $.

For any $r <1$ we can find $r < \frac {u_n}{u_{n-1}} < 1$. So we can not find an appropriate $r <1$.

So we failed the hypothesis. The ratio test fails.