Determine whether $a_n$ is convergent Let $a_n = \frac{2n}{3n+1}$
(a) Determine whether $\{a_n\}$ is convergent.
(b) Determine whether $\sum_{n=1}^{\infty}a_n$ is convergent.
I am having trouble getting started with (a). Is the question asking to find the limit of $a_n$? If so that would be $\frac{2}{3}$ I believe.
Then for (b), I am trying to convert this into a geometric series, but I haven't been able to figure out how to. I observe that 
$\begin{align}
a_1 = \frac{2}{4}, a_2 = \frac{4}{7}, a_3 = \frac{6}{10}
\end{align}$
but I don't see a pattern.
 A: You are right about the limit of the sequence $a_n=\frac{2n}{3n+1}$.
We have indeed that $\lim_{n\to\infty} a_n=\frac23$
There are several ways to show this. If you know about limit theorems, you can use them like this:
It is $\frac{2n}{3n+1}=\frac{2}{3+\frac1n}$. Now the numerator coverges obviously against $2$ and the denominator converges against $3$. So the fraction converges against $\frac23$.
An other would be straight by definition:
$\forall\varepsilon >0\exists N\in\mathbb{N}$ sucht that for every $n\geq N$ holds that $|a_n-\frac23|<\varepsilon$.
'Sketch':
Let $\varepsilon >0$ be arbitrary. We have to find $N$.
It is $|a_n-\frac23|=|\frac{2n}{3n+1}-\frac23|=|\frac{6n-6n-2}{3(3n+1)}|=\frac{2}{3(3n+1)}<\frac{1}{3n+1}<\frac1n<\frac1N$.
If we choose $N=\lceil \frac1\varepsilon\rceil$ we could conclude the proof.
For b):
As mentioned in the comments a series can only converge if you take the sum over a null sequence.
If the sequence is not a null sequence, the series can not converge.
However if you have a null sequence it is not sure that the series converges also.
The standard counterexample is:
$\sum_{k=1}^\infty \frac1k$ 
the harmonic series, which does not converge!
