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?
 A: 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 $\sum_{n=1}^{\infty}a_{n}$
A: the reason is that $\frac1x$ doesnt converge to the limit fast enough.
yes $\lim\limits_{x\to\infty}\frac1x=0$ but notice that, for example, that also $\lim\limits_{x\to\infty}\frac1{x^2}=0$, what is the difference between the $2$?
$x^{-1}>x^{-2}$ for $x>1$. what does it means? well $x^{-2}$ goes to $0$ as $x$ goes to $\infty$ faster than $x^{-1}$. @B2VSi showed the proof of why $x^{-1}$ doesnt converges so i wont add it here.
another examplpe that can make things clearer is this: $\sqrt{x^4+x^2}>x^2$ as x goes to infinity, it is clear. and  $\sqrt{x^4+x^2}-x^2=\frac12$ as x goes to infinity.we also have $\sqrt{x^4+x^3}-x^2=\infty$ as x goes to infinity. now this is normal, after all $\sqrt{x^4+x^2}-x^2<\sqrt{x^4+x^3}-x^2$ as x goes to infinity. but what really changed? after all both of the are in the form of $\infty-\infty$, all 3 of the expressions $\left(\sqrt{x^4+x^3}\,,\sqrt{x^4+x^2}\,,x^2\right)$ doesnt converge. what changed is the 'speed' they goes to infinity, $\sqrt{x^4+x^3}$ goes to infinity faster then $\sqrt{x^4+x^2}$, so fast that it is infinitely bigger than $x^2$ at $\lim\limits_{x\to\infty}$ while $\sqrt{x^4+x^2}$ was only $\frac12$ more.
so to conclude what matters in series and in convergences at all is the rate of the change of the expressions and not if one value converge or not  
