Note that the series is alternating. To be able to use the ratio test, we will have to compute the ratio:

\(\displaystyle \left|\frac{a_{n+1}}{a_n}\ \right|=\frac{(-1)^{n+2}}{n+1}\cdot \frac{n}{(-1)^{n+1}}=\frac{n}{n+1}\)

Now we need to see that :

\(\displaystyle \lim_{n\rightarrow \infty }\left|\frac{a_{n+1}}{a_n}\ \right|=\lim_{n\rightarrow \infty}\frac{n}{n+1}=1\).

Since \(\displaystyle \lim_{n\rightarrow \infty }\left|\frac{a_{n+1}}{a_n}\ \right|=1\) we can't conclude by using the ratio test. We will have to call upon another test to show that the series is convergent.

To show that the statement above is false, we will consider the following example.

consider the series \(\displaystyle \sum_{n=1}^{\infty}a_{n}\) where \(\displaystyle a_{n}\) is given by:

\(\displaystyle a_{n}=\frac{1}{n(n+1)}\) clearly the series has positive terms. Furthermore, we have

\(\displaystyle \lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n}=\frac{1}{(n+1)(n+2)}{\cdot \frac{n(n+1)}{1}}=\frac{n}{n+2}=1\), meaning the series can be either convergent or divergent.

However, the series :\(\displaystyle \sum_{n=1}^{\infty}a_{n}\) is convergent (use for example the integral test to see that it is convergent).

Therefore, the original statement is false.

Note that the series is positive.

As it is required we will use the ratio test to check for the nature of the series. 

We have \(\displaystyle \frac{a_{n+1}}{a_n}=\frac{a^{n+1}}{(n+1)^b}\frac{n^b}{a^n}=a\left(\frac{n}{n+1}\right)^b\).

Therefore, \(\displaystyle \lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\lim_{n\rightarrow\infty} a(\frac{n}{n+1})b\)\(\displaystyle \lim_{n\rightarrow\infty}\frac{n^b}{(n+1)^b}=a(\lim_{n\rightarrow}\frac{n}{n+1})^b=a\)

\(\displaystyle L=\lim_{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|\) if L>1 the series diverges, if L<1 the series converges absolutely, and if L=1 the series may either converge or diverge.

Since \(\displaystyle a< \frac{1}{999}< 1\) the ratio test concludes that the series converges absolutely.

Determine the convergence of the series based on the limits.

\(\displaystyle \newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1\)

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take \(\displaystyle \lim_{n\rightarrow \infty}\).

\(\displaystyle \frac{\frac{1}{5^{n+1}}}{\frac{1}{5^n}}\)\(\displaystyle = \frac{5^n}{5^{n+1}}=0.2\)

\(\displaystyle \lim_{n\rightarrow \infty} 0.2 \rightarrow 0.2\)

Determine the convergence of the series based on the limits.

\(\displaystyle \newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1\)

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take \(\displaystyle \lim_{n\rightarrow \infty}\)

\(\displaystyle \frac{\frac{1}{2^{n+1}}}{\frac{1}{2^n}}\)\(\displaystyle = \frac{2^n}{2^{n+1}}=0.5\)

\(\displaystyle = \lim_{n\rightarrow \infty} 0.5 \rightarrow 0.5\)

Determine the convergence of the series based on the limits.

\(\displaystyle \newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1\)

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take \(\displaystyle \lim_{n\rightarrow \infty}\).

\(\displaystyle \frac{\frac{e^{n+1}}{3^{n+1}}}{\frac{e^n}{3^n}}\)\(\displaystyle = \frac{\frac{e^{n+1}}{e^n}}{\frac{3^{n+1}}{3^(n)}}\) \(\displaystyle = \frac{e}^{3}\) 

\(\displaystyle \lim_{n\rightarrow \infty} \frac{e}{3} \rightarrow \frac{e}{3}\)

Determine the convergence of the series based on the limits.

\(\displaystyle \newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1\)

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take \(\displaystyle \lim_{n\rightarrow \infty}\).

\(\displaystyle \frac{\frac{3^{n+1}}{2^{n+1}}}{\frac{3^n}{2^n}}\) \(\displaystyle = \frac{\frac{3^{n+1}}{3^n}}{\frac{2^{n+1}}{2^n}}=\frac{3}{2}\)

\(\displaystyle \frac{3}{2}= 1.5\)

\(\displaystyle =\lim_{n\rightarrow \infty} 1.5 \rightarrow 1.5\)

Determine the convergence of the series based on the limits.

\(\displaystyle \newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1\)

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take \(\displaystyle \lim_{n\rightarrow \infty}\).

\(\displaystyle \frac{\frac{e^{n+1}}{5^{n+1}}}{\frac{e^n}{5^n}}\)\(\displaystyle = \frac{\frac{e^{n+1}}{e^n}}{\frac{5^{n+1}}{5^n}}\)\(\displaystyle = \frac{e}^{5}\)

\(\displaystyle \lim_{n\rightarrow \infty} \frac{e}{5} \rightarrow \frac{e}{5}\)

Determine the convergence of the series based on the limits.

\(\displaystyle \newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1\)

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take \(\displaystyle \lim_{n\rightarrow \infty}\).

\(\displaystyle \frac{\frac{e^{n+1}}{4^{n+1}}}{\frac{e^n}{4^n}}\) \(\displaystyle = \frac{\frac{e^{n+1}}{e^n}}{\frac{4^{n+1}}{4^n}}=\frac{e}{4}\) \(\displaystyle ,\lim_{n\rightarrow \infty} \frac{e}{4} \rightarrow\) \(\displaystyle \frac{e}^{4}\)

Determine the convergence of the series based on the limits.

\(\displaystyle \newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1\)

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take \(\displaystyle \lim_{n\rightarrow \infty}\).

\(\displaystyle \frac{\frac{(n+1)!}{e^{n+1}}}{\frac{n!}{e^n}}\) \(\displaystyle = \frac{\frac{(n+1)!}{n!}}{\frac{e^{n+1}}{e^n}}\) \(\displaystyle = \frac{n+1}^{e}\) \(\displaystyle ,\lim_{n\rightarrow \infty} \frac{n+1}{e} \rightarrow \infty\)