Infinite series Let $a_n = \frac{2n}{3n+1}$
a) Determine whether ${a_n}$ is convergent.
I got that it was convergent by taking the $\lim_{n\to \infty}$ $\frac{2n}{3n+1}$ and multiplying the numerator and denominator by $\frac{1}{n}$ and obtaining $\lim_{n \to \infty}$ $\frac{2}{3+\frac{1}{n}}$ $=$ $\frac{2}{3}$ and therefore converges. 
b) determine whether $\sum_{n=1}^\infty$ $a_n$ is convergent 
I got the question wrong. I used my work as above to say that the series converges to $\frac{2}{3}$ but the book says diverges and I think that is because by theorem 7, test for divergence which states that if $\lim_{n \to \infty}a_n\neq 0$ or does not exist then the series $\sum_{n=1}^\infty$ $a_n$ is divergent. By my work in part a I showed that the $\lim_{n \to \infty}$ $a_n$ exists and $\neq$ $0$ and is therefore divergent. Is that correct? 
Important:
I have a hard time understanding conceptually what the difference is between part a and b. How the limit can in one instance indicate convergence but in an another instance with different notation indicate divergence. What is the difference between those two notations? For what types of series would the convergence test not work?
 A: The first part is totally fine. I am taking the liberty to assume that you are aware about the definition of convergence of a sequence in terms of $\epsilon$ and $N$.
When considering part (b), you should keep in mind that in order the answer the question whether $\sum_{n=0}^{\infty}a_n$ is convergent you actually need to answer whether the sequence $(s_k)$ of partial sums defined as
$$
s_k=\sum_{n=1}^{k}a_n
$$
is convergent or not. This is required because $(s_k)$ is a sequence of real numbers (or complex numbers) because we know how to add a finite number of real numbers (or complex numbers) but can't be sure what an infinite sum would be unless we pass to the limiting procedure. 
Here is a definition that actually shows how a series is different from a sequence. 
${\bf Definition}$: A series of real numbers is a pair $(\{a_n\},\{s_k\})$, where $\{a_n\}$ and $\{s_k\}$ are sequences of real numbers such that
 $$
s_k=\sum_{n=1}^{k}a_n.
$$ 
$\textbf{ In short hand notation one writes}$ $ \sum_{n=1}^{\infty}a_n $$\textbf {for a series}$. And a series $ \sum_{n=1}^{\infty}a_n $ is said to be convergent if the sequence $\{s_k\}$, called the sequence of partial sums, is convergent.
And as far as tests for convergence of series are concerned, any test can be applied to any series though it is an entirely different issue to choose a suitable test that might settle down the question.
A: There is a difference between a sequence, and a series which is the sum of the terms of a sequence.  An easy way to understand this is to look at the geometric sequence and series that you saw in high school:  
Let $a_n = rx^n$.  For example, with $r=1$ and $x=\frac12$, 
$$
a_0=1, a_1=\frac12, a_2=\frac14, a_3=\frac18 \cdots
$$
That is a sequence and it is easy to see that if $|x|<1$ then as $n\to\infty$ the value of the $n$-th term of the sequence gets smaller and smaller and goes to zero.
Now look at the related series $\sum_n a_n$.  This is a bit of shorthand for looking at sums of the form
$$
s_k = \sum_{n=0}^{k} a_n
$$ 
For the geometric series you learned that (for the case $a_n = x_0 x^n$)
$$
\sum_{n=0}^{k} x_0 x^n = x_0\frac{1-x^{k+1}}{1-x}
$$
(and you may have noticed that this formula breaks down if $x=1$).  At any rate, for our example, 
$$
\sum_{n=0}^{k} \frac{1}{2^n} = \frac{1-2^{k+1}}{1-\frac12} = 2- \frac{1}{2^k}
$$
and you can see those partial sums, for example, 
$$ 1 + \frac12 + \frac14 + \frac18 = 1\frac78$$
Showing that $\lim_{n\to\infty} = 0$ is a necessary condition for $\sum a_n$ to converge -- look what happens to the geometric series if $x>1$. But it is not sufficient.  For example, look at $a_n = \frac1n$ so that 
$$
\sum_{n=1}^\infty a_n = 1+\frac12 + \left(\frac13 + \frac14\right) + \left(\frac15 +\frac16 + \frac17+ \frac18\right) + \cdots 
\\ > 1+\frac12 + \left(\frac14 + \frac14\right) + \left(\frac18 +\frac18 + \frac18+ \frac18\right) + \cdots = 1+\frac12+\frac12+\frac12+\cdots
$$
This demonstrates that $\sum_{n=1}^\infty a_n$ "is infinity", that is, it does not converge.  We need some additional other condition to be sure that a series converges.
A simple test that often works is the ratio test:
$$
\lim_{n\to\infty} \frac{|a_{n+1}|}{|a_{n}|} <1 \implies \sum a_n \mbox{ converges} \\ \lim_{n\to\infty} \frac{|a_{n+1}|}{|a_{n}|} >1 \implies \sum a_n \mbox{ diverges} 
$$
When that limit is equal to $1$, then the ratio test has nothing to say. FOr your problem, the ratio test comes down to 
$$
\lim_{n\to\infty} \frac{6n^2+8n+2}{6n^2+8n} = 1
$$
So how can you decide whether the series converges?
Well, you might notice that for all $n\ge 1$ we have  $\frac{2n}{3n+1} \ge \frac12$
So the series sum is greater than $\frac12+\frac12+\frac12+\cdots$ so it must diverge.
Or you can use the fact that the limit of $a_n$ exists and is $\frac23$ which is non-zero; as your book states, when this happens the series (that is, the sum of $a_n$) cannot possibly converge.
