The series in question is:

$\sum_{n=1}^{\infty} \frac{1}{n^{2p} + n}$

To determine for which values of $p$ the series diverges, let's first analyze the behavior of the general term.

For large values of $n$, the term $n^{2p} + n$ behaves approximately as $n^{2p}$, since the term $n^{2p}$ dominates over $n$ for large $n$. Thus, the general term behaves like:

$\frac{1}{n^{2p}}$

Now, we can use the p-series test to determine convergence or divergence. A p-series of the form $\sum \frac{1}{n^q}$ converges if $q > 1$ and diverges if $q \leq 1$.

Therefore, we require $2p > 1$ for convergence. Solving for $p$, we get:

$2p > 1 \quad \Rightarrow \quad p > \frac{1}{2}$

Hence, the series converges when $p > \frac{1}{2}$ and diverges when $p \leq \frac{1}{2}$.

Conclusion:

The series diverges for $p \leq \frac{1}{2}$, which corresponds to option A.

Let me know if you'd like further explanation or details! Here are some follow-up questions:

1. How does the comparison test relate to this series?
2. Can we apply the integral test to this series?
3. What is the behavior of the series for values of $p$ much larger than $1/2$?
4. How does the term $n^{2p} + n$ affect the convergence for small $n$?
5. What other types of series tests could we apply to analyze convergence here?

Tip: For series with terms involving polynomials like this, it's helpful to compare the terms with simpler p-series forms to determine convergence behavior.