The Formula

$$\sum\limits_{n=1}^{\infty}a_{n} \text{ diverges if } \lim\limits_{n\rightarrow\infty}a_{n} \text{ fails to exists or does not equal zero.}$$
Note: If the limit equals zero that does not mean that the series converges . It means that it could converge. If the limit equals zero, more tests have to be performed to prove whether or not the series converges. However, if the limit is not zero, we know the series diverges. This is why the test is also known as the divergence test.
The Nth term test is often the first test that you should use to rule out the series diverges before you go any further.
 Series Limit Conclusion $$\sum\limits_{n=0}^{\infty}\frac{n}{2n+1}$$ $$\lim\limits_{n\rightarrow\infty}\frac{n}{2n+1}=\frac{1}{2}$$ diverges since $$\lim\limits_{n\rightarrow\infty}\frac{n}{2n+1}\neq 0$$ $$\sum\limits_{n=0}^{\infty}\frac{n!+1}{3n!-5}$$ $$\lim\limits_{n\rightarrow\infty}\frac{n!+1}{3n!-5}=\frac{1}{3}$$ diverges since $$\lim\limits_{n\rightarrow\infty}\frac{n!+1}{3n!-5}\neq0$$ $$\sum\limits_{n=0}^{\infty}\frac{n}{ln(n)}$$ $$\lim\limits_{n\rightarrow\infty}\frac{n}{ln(n)}=\infty$$ L'Hopitals Rule diverges since $$\sum\limits_{n=0}^{\infty}\frac{n}{ln(n)}$$ $$\sum\limits_{n=0}^{\infty}n \cdot sin(\frac{1}{n})$$ $$\lim\limits_{n\rightarrow\infty}n \cdot sin(\frac{1}{n})=1$$ L'Hopitals Rule diverges since $$\lim\limits_{n\rightarrow\infty}n \cdot sin(\frac{1}{n})\neq0$$ $$\sum\limits_{n=0}^{\infty}\frac{1}{n}$$ $$\lim\limits_{n\rightarrow\infty}\frac{1}{n}=0$$ could possibly converge
In fact, in the last example we found that $$\lim\limits_{n\rightarrow\infty}\frac{1}{n}=0$$, however, we know that the series $$\sum\limits_{n=0}^{\infty}\frac{1}{n}$$ is a P-series that diverges.
On the other hand, consider the series $$\sum\limits_{n=0}^{\infty}\frac{1}{n^2}$$ the $$\lim\limits_{n\rightarrow\infty}\frac{1}{n^2}=0$$ . In this case, we have a P-series with p=2 so this series converges.