Why is the limit of $\frac{1}{n}$ is $0$ however the series $\sum_{n=1}^{\infty} \frac{1}{n}$ is divergent? As I recall, according to the test for divergence, if you have a series $\sum_{n=1}^{\infty}a_{n}$ and if the limit of $a_{n}$ is $0$, then the series is convergent. the limit of $\frac{1}{n}$ is $0$. However, if we use the integral test, we realize that the series of $\frac{1}{n}$. I am confused to why it is like this. Am I overlooking something?

Actually, if $a_{n}$ does not tend to $0$, then $\sum_{n=1}^{\infty}a_{n}$ is divergent. The opposite is not true.

A good example is $a_n = 1/n$. To visualize this, consider the series,
$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + ...$$
Now replace all $a_n$ to the closest power of $2$ which is smaller. For example $\frac{1}{3}$ becomes $\frac{1}{4}$. The new series is then,
$$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{16} + \frac{1}{16} + ...$$
Notice that if I group the terms, what happens
$$1 + \frac{1}{2} + (\frac{1}{4} + \frac{1}{4}) + (\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}) + (\frac{1}{16} + \frac{1}{16} + ...$$
Each of the terms add up to half
$$1 + \frac{1}{2} + (\frac{1}{2} ) + (\frac{1}{2}) + (\frac{1}{2} ) + ...$$
From here I'm pretty sure you can see that this series diverges. Now coming to the original series. Since the new series corresponding terms are smaller and this series diverges, so does the series $\sum_{n=1}^{\infty}a_{n}$
