Finding the sum of the series with $a_n = \frac{4n}{6n+7}$ 
In my calc 2 course, one of the homework problems is to let
$$a_n = \frac{4n}{6n+7}$$
and then find the limit of its sequence and the sum of its series.

I started with the limit:
$$\lim_{n\to \infty} \frac{4n}{6n+7}$$
Using L'Hopital, I ended up with $\dfrac{2}{3}$ as the answer.
The sum of the series is where I seem to be having trouble.
So with $\sum_{n=1}^\infty \frac{4n}{6n+7}$, I generated the first 3 terms of the series:
$$\frac{4}{13} + \frac{8}{19} + \frac{12}{25} + \cdots $$
To me this series does not appear to be geometric, so I tried to utilize a definition I was provided with in class:

A series converges iff the sequence, $S_n$ converges. If $S_n$ converges, we say that $\lim_{n\to \infty} S_n = S = \sum_{n=1}^\infty \frac{4n}{6n+7}$, where $S$ is some real finite number.

Based on this, the sum of the series would have to also be $\frac{2}{3}$ but when I entered this as my answer, it was marked as incorrect.
So at this point, I'm thinking I might have gotten this definition incorrect or there is some way to algebraically manipulate $a_n$ to get a geometric "form" that I'm not seeing.
 A: There are already several complete answers here. I would like to suggest a strategy.
Spend a little time just thinking about the shape of the problem before you start applying tools.
When $n$ is large, the fraction
$$
\frac{4n}{6n+7}
$$
is very close to
$$
\frac{4n}{6n } = \frac{2}{3}
$$
(since the $+7$ in the denominator is negligible)
so that will be the limit of the sequence. You don't need anything like L'Hopital's rule to prove that. I would accept a student's statement that it was "obvious" if the surrounding work was good. If you do want a more formal proof, write
$$
\frac{4n}{6n+7} =
\frac{4}{6+7/n}
$$
and look at the limit of the numerator over the limit of the denominator.
Once you know the terms are near $2/3$ when $n$ is large you know the sequence ehds up looking like
$$
\cdots + 2/3 + 2/3 + \cdots
$$
which clearly diverges. The general term does not have limit $0$.
A: You obviously confused the limit of the sequence $a_n $ with that of its partial sums:
$$S_n=\sum_{k=1}^n a_k,
$$
which diverges in this case because $\lim_{n\to\infty} a_n=\frac23\ne0$.
A: Since you seem to be unfamiliar with the term test for series divergence I am writing a more detailed answer. Let $(a_n)_{n\geq 1}$ be a sequence of real numbers. We define the sequence of partial sums for $(a_n)$ as follows:
$$s_n = a_1 + a_2 + \dots + a_n, \;\; n\geq 1$$
We say that the series with general term $a_n$ (or $\sum a_n$) converges if the sequence $(s_n)$ converges to a real number $l$. A necessary condition for a series to converge is that $a_n \to 0$. We can see this by noting that
$$a_n = s_n - s_{n-1}, \;\; n \geq 2$$
Assuming that $s_n \to l$ we can take limits on both sides of the above equality and conclude that $a_n \to l - l = 0$ since $(s_{n-1})$ converges to $l$ as well.
In your case, you have shown that $a_n \not\to 0$ so by contraposition the series with general term $a_n$ cannot converge.
