How do I show this series diverges? The series in question is $$\sum_{n=1}^\infty\frac{n^2+1}{2n^2+5}$$
If $a_n$ is the $n$'th term of this sum, then $a_n \rightarrow \frac{1}{2}$ as $n\rightarrow \infty$. I believe this implies that the series doesn't converge - since this is effectively the negation of the statement "If a series converges, then $a_n$ tends to $0$ as $n$ tends to $\infty$". If I prove that the sequence $x_n=\frac{n^2 + 1}{2n^2 +5}$ converges to $\frac{1}{2}$, have I then proven that the series diverges? Or should I apply a convergence test and show that it doesn't pass one?

Just for your curiosity.
As already said, the problem is over as soon as you showed that the term tends to anything which is not zero.
If fact, we can even approximate the partial sums
$$\frac{n^2+1}{2n^2+5}=\frac{1}{2}-\frac{3}{4 \left( n^2+\frac52\right)}$$
Now, use partial fraction decomposition
$$\frac{1}{ n^2+\frac52}=\frac 1{(n+i\sqrt{\frac 52})(n-i\sqrt{\frac 52})}=\frac 1 {2a}\left(\frac 1{n-a} -\frac 1{n+a} \right)$$
$$\sum_{n=1}^p\frac 1{n+a} = H_{p+a}-H_a\qquad \text{and}\qquad \sum_{n=1}^p\frac 1{n-a} = H_{p-a}-H_{-a}$$ and using the asymptotics
$$H_q=\gamma +\log \left({q}\right)+\frac{1}{2 q}-\frac{1}{12
   q^2}+O\left(\frac{1}{q^3}\right)$$ After some minor simplifications of the complex numbers, you should end with
$$S_p=\sum_{n=1}^p\frac{n^2+1}{2n^2+5}=\frac{p}{2}+\frac{3}{40} \left(2- \sqrt{10}\, \pi  \,\coth \left(\sqrt{\frac{5}{2}} \pi
   \right)\right)+\frac{3}{4 p}-\frac{3}{8 p^2}+O\left(\frac{1}{p^3}\right)$$ Try it for $p=10$; the exact value is 
$$S_{10}=\frac{255436901298463}{57072223853645}\approx 4.47568 $$ while the truncated expansion given above leads to $4.47608$.
