No the series does not converge. The given problem is the harmonic series, which diverges to infinity.

The series converges. The given problem is the alternating harmonic series, which converges by the alternating series test.

Only the Integral Test will work on the Harmonic Series, \(\displaystyle \sum_{n=1}^\infty \frac{1}{n}\).

To use the Integral Test, we evaluate

\(\displaystyle \int_1^\infty \frac{1}{x}dx = [\ln(x)]_1^\infty = \lim_{x \to \infty}\ln(x)-0 =\infty\), which shows that the series diverges.

Since \(\displaystyle \frac{1}{n} \rightarrow 0\), the Limit Test for Divergence fails.

The Ratio Test and the Root Test will always yield the same conclusion, so if one test fails, the both fail and vise versa.

For the Ratio Test,

\(\displaystyle \lim_{x \to \infty}|\frac{1}{n+1}\times \frac{n}{1}| = \lim_{x \to \infty}\frac{n}{n+1} = 1\). Since the result of the limit is \(\displaystyle 1\), both tests fail.

The given series is called generalized harmonic series.

The series converges, if \(\displaystyle p>1\), and diverges, if \(\displaystyle p\leq1\).

This type of series we can frequently check for convergence/divergence using the Alternating Series Test.

\(\displaystyle \sum_{n=1}^\infty(-1)^{n+1}a_n=a_1-a_2+a_3-a_4+...\)

The terms with an odd value for \(\displaystyle n\) become negative since \(\displaystyle (-1)^n=-1\) and the terms with an even value for \(\displaystyle n\) are positive. This creates the alternating signs to occur within the sum.

To differentiate the function we will need to use the Power Rule which states:

\(\displaystyle f(x)=ax^n \rightarrow f'(x)=nax^{n-1}\)

Looking at our function we can first simplify the equation.

\(\displaystyle f(x)=x^4+\frac{x}{3x^2}+9=x^4+\frac{1}{3x}+9\)

Applying the Power Rule we get:

\(\displaystyle f'(x)=4x^3-\frac{1}{3x^2}\)

The series converges conditionally.

The absolute values of the series \(\displaystyle \sum_{n=0}^{\infty }\left |\frac{(1)^n}{n} \right |\) is a divergent p-series with \(\displaystyle p\leq 1\).

However, the the limit of the sequence \(\displaystyle \lim_{n\rightarrow \infty }\frac{(-1)^n}{n}=0\) and it is a decreasing sequence.

Therefore, by the alternating series test, the series converges conditionally.    

Using the root test, 

\(\displaystyle \lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{3^{n+1} x^{n+1}}{(n+1)!} \cdot \frac{n!}{3^nx^n} \right |=\left | 3x\right |\lim_{n\rightarrow \infty }\left | \frac{1}{n+1} \right |=0\)

Because 0 is always less than 1, the root test shows that the series converges for any value of x. 

Therefore, the interval of convergence is:

\(\displaystyle (-\infty, \infty)\)

We can use the alternating series test to show that

\(\displaystyle \small \sum_{n=1}^{\infty} (-1)^n \sin\frac{1}{n}\)

converges.

We must have \(\displaystyle \small \sin \frac{1}{n}\geq 0\)  for \(\displaystyle \small n\geq 1\) in order to use this test. This is easy to see because \(\displaystyle \small \frac{1}{n}\) is in\(\displaystyle \small \small \left[0,\frac{\pi}{2}\right]\) for all \(\displaystyle \small n\geq 1\) (the values of this sequence are \(\displaystyle \small 1,1/2,1/3,1/4,...\)), and sine is always nonzero whenever sine's argument is in \(\displaystyle \small \small \left[0,\frac{\pi}{2}\right]\).

Now we must show that

1. \(\displaystyle \small \small \lim_{n\to\infty} \sin \frac{1}{n}=0\)

2. \(\displaystyle \small \sin \frac{1}{n}\) is a decreasing sequence.

The limit 

\(\displaystyle \small \lim_{n\to\infty}\frac{1}{n}=0\)

implies that 

\(\displaystyle \small \small \small \lim_{n\to\infty} \sin \frac{1}{n}=\sin 0=0\)

so the first condition is satisfied.

We can show that \(\displaystyle \small \small \small \sin \frac{1}{n}\) is decreasing by taking its derivative and showing that it is less than \(\displaystyle \small 0\) for \(\displaystyle \small n\geq 1\):

\(\displaystyle \small \small \small \small \small \frac{d}{dn}\sin \frac{1}{n}=-\frac{1}{n^2}\cos\frac{1}{n}\)

The derivative is less than \(\displaystyle \small 0\), because \(\displaystyle \small -\frac{1}{n^2}\) is always less than \(\displaystyle \small 0\), and that \(\displaystyle \small \cos \frac{1}{n}\) is positive for \(\displaystyle \small n\geq 1\), using a similar argument we used to prove that \(\displaystyle \small \small \sin \frac{1}{n}\geq 0\) for \(\displaystyle \small n\geq 1\). Since the derivative is less than \(\displaystyle \small 0\), \(\displaystyle \small \small \small \sin \frac{1}{n}\) is a decreasing sequence. Now we have shown that the two conditions are satisfied, so we have proven that 

\(\displaystyle \small \sum_{n=1}^{\infty} (-1)^n \sin\frac{1}{n}\)

converges, by the alternating series test.

The series given is an alternating series.  

Write the three rules that are used to satisfy convergence in an alternating series test.

For \(\displaystyle \sum_{n=1}^{\infty} (-1)^{n+1} x_n= x_1-x_2+x_3-x_4+...\):

\(\displaystyle \textup{1. All the } x_n \textup{ terms are positive.}\)

\(\displaystyle \textup{2. The terms must be decreasing: } x_n \geq x_{n+1}\)

\(\displaystyle \textup{3. The limit }\lim_{n \to \infty}x_n=0.\)

\(\displaystyle \sum_{n=0}^{\infty} (-1)^n (\frac{n^n}{8^n})=\sum_{n=0}^{\infty} (-1)^n (\frac{n}{8})^n\)

The first and second conditions are satisfied since the terms are positive and are decreasing after each term.

However, the third condition is not valid since \(\displaystyle \lim_{n \to \infty}x_n\neq 0\) and instead approaches infinity.

The correct answer is:

\(\displaystyle \textup{Diverge, since}\lim_{n \to \infty}x_n\neq 0.\)