Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{2} } }\) converges or diverges.

Let's try the Divergence Test.
\(\displaystyle{ \lim_{n \to \infty}{n/2} = \frac{\infty}{2} = \infty }\)
Since the limit is not zero, the series diverges by the Divergence Test.

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{2} } }\) diverges by the Divergence Test.

Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5n}{n^3} } }\) converges or diverges.

Starting with the divergence test, we need to determine the limit \(\displaystyle{ \lim_{n \to \infty}{ \frac{5n}{n^3} } }\).
There are two main ways to do this.
1. The first, and easiest way, is to use algebra.
2. The second way is to use L'Hôpital's Rule.
We will use algebra by multiplying the numerator and denominator by \(1/n\).
\(\begin{array}{rcl} \displaystyle{ \lim_{n \to \infty}{ \frac{5n}{n^3} }} & = & \displaystyle{ \lim_{n \to \infty}{ \frac{5n}{n^3} \frac{1/n}{1/n} }} \\ & = & \displaystyle{ \lim_{n \to \infty}{ \frac{5}{n^2}}} = 0 \end{array}\)
Since the limit is zero, we cannot say anything about whether this series converges or diverges using only the divergence test.
If you have not studied the p-series yet, skip the next section and jump to the notes.
p-series test - By canceling the \(n\) in the numerator with one in the denominator (which we do since the canceling operation does not change the domain), we have \(a_n = 5/n^2\). This is a p-series with \(p=2>1\). So this series converges by the p-series test.
Notes
1. Even though the divergence test cannot be used to determine convergence or divergence, we didn't just stop working the problem. The problem statement said to determine convergence or divergence. So stopping after applying the divergence test would not be the completion of the problem. We actually needed to determine convergence or divergence. So once the first test failed, we moved on to another test until one of the tests determined convergence or divergence.
2. If you have not studied the p-series test yet, do not despair. Just make sure you understand that you cannot determine convergence/divergence from the divergence test when the limit is zero and that you can't just stop working the problem when the divergence test doesn't work.

Convergence/divergence cannot be determined by the divergence test. The series converges by the p-series test.

Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^2+3n+1}{2n^2+4} } }\) converges or diverges.

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^2+3n+1}{2n^2+4} } }\) diverges by the Divergence Test.

Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^2+2}{n^2+1} } }\) converges or diverges.

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^2+2}{n^2+1} } }\) diverges by the Divergence Test.

Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \cos\left[ \frac{1}{n^2} \right] } }\) converges or diverges.

Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \ln(\cos n) } }\) converges or diverges.

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \ln(\cos n) } }\) diverges by the Divergence Test.

Determine whether the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{n~\cos(n\pi)}{2n-1}}}\) converges or diverges.

\(\displaystyle{ a_n = \frac{n~\cos(n\pi)}{2n-1} }\)

\(\displaystyle{\lim_{n \to \infty}{ \frac{n~\cos(n\pi)}{2n-1} } = \lim_{n \to \infty}{ \frac{\cos(n\pi)}{2-1/n} } }\)

In this limit, we know that \(\displaystyle{ \lim_{n \to \infty}{1/n} = 0 }\), leaving \(2\) in the denominator. In the numerator, \(\cos(n\pi)\) oscillates between \(-1\) and \(1\). So we are left with the series oscillating between \(-1/2\) and \(1/2\) and therefore, the limit \(\displaystyle{\lim_{n \to \infty}{ \frac{n~\cos(n\pi)}{2n-1} } \neq 0 }\)
So, by the divergence test, the series diverges.
The divergence test is the best test for this series. For a discussion on using other tests, see the practice problem page dedicated to this problem.

The series diverges by the divergence test.

Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \sin(n\pi/2) } }\) converges or diverges.

In order to determine what the limit \(\displaystyle{\lim_{n\to\infty}{\sin(\pi n/2)}}\) does, you need to think about the sine function. Basically, the sine function takes angles and maps them to the range \(-1 \to 1\). So as \(n \to \infty\) the sine function oscillates between -1 and 1. It never settles at a specific value and, more importantly for us here, it never settles or approaches zero.
Conclusion: Since \(\displaystyle{\lim_{n\to\infty}{\sin(\pi n/2)}\neq 0}\), this series diverges by the Divergence Test.

The series \(\displaystyle{\sum_{n=1}^{\infty}{\sin(n\pi/2)}}\) diverges by the Divergence Test.