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?

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  
