We cannot test for convergence of a \(\displaystyle p\)-series using the ratio test. Observe,

For the series \(\displaystyle \sum_{n=1}^\infty \frac{1}{n^p}\),

\(\displaystyle \lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} | \frac{n^p}{(n+1)^p}| = \lim_{n \to \infty} (\frac{n}{n+1})^p = 1^p =1\).

Since this limit is \(\displaystyle 1\) regardless of the value for \(\displaystyle p\), the ratio test is inconclusive.

In order to figure out if 

\(\displaystyle \sum_{n=1}^{\infty} \frac{(-9)^n}{5^{3n+2}(n+1)}\)

is divergent, convergent or neither, we need to use the ratio test.

Remember that the ratio test is as follows.

Suppose we have a series \(\displaystyle \sum a_n\). We define,

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

Then if 

\(\displaystyle L< 1\), the series is absolutely convergent.

\(\displaystyle L>1\), the series is divergent.

\(\displaystyle L=1\), the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply the ratio test to our problem.

Let  

\(\displaystyle a_n=\frac{(-9)^n}{5^{3n+2}(n+1)}\)

and

\(\displaystyle a_{n+1}=\frac{(-9)^{n+1}}{5^{3(n+1)+2}((n+1)+1)}\)

\(\displaystyle =\frac{(-9)^{n+1}}{5^{3n+5}(n+2)}\)

Now 

\(\displaystyle L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(-9)^{n+1}}{5^{3n+5}(n+2)}}{\frac{(-9)^n}{5^{3n+2}(n+1)}} \right |\)

\(\displaystyle =\lim_{n\rightarrow \infty}\left | \frac{(-9)^{n+1}5^{3n+2}(n+1)}{(-9)^n5^{3n+5}(n+2)} \right |\)

Now lets simplify this expression to 

\(\displaystyle =\lim_{n\rightarrow \infty}\left | \frac{(-9)^{1}(n+1)}{5^{3}(n+2)} \right |\)

\(\displaystyle =\frac{9}{125}\lim_{n\rightarrow \infty}\left | \frac{n+1}{n+2} \right |\)

\(\displaystyle =\frac{9}{125}*1=\frac{9}{125}\).

Since 

\(\displaystyle \frac{9}{125}< 1\).

We have sufficient evidence to conclude that the series is convergent.

In order to figure if 

\(\displaystyle \sum_{n=1}^{\infty} \frac{n!}{2^{n+1}}\)

is convergent, divergent or neither, we need to use the ratio test.

Remember that the ratio test is as follows.

Suppose we have a series \(\displaystyle \sum a_n\). We define,

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

Then if 

\(\displaystyle L< 1\), the series is absolutely convergent.

\(\displaystyle L>1\), the series is divergent.

\(\displaystyle L=1\), the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply the ratio test to our problem.

Let  

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

and

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

Now 

\(\displaystyle L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)!}{2^{n+2}}}{\frac{n!}{2^{n+1}}} \right |\)

\(\displaystyle =\lim_{n\rightarrow \infty}\left | \frac{2^{n+1}(n+1)!}{2^{n+2}n!} \right |\).

Now lets simplify this expression to 

\(\displaystyle =\lim_{n\rightarrow \infty}\left | \frac{n+1}{2} \right |\)

\(\displaystyle =\frac{1}{2}\lim_{n\rightarrow \infty}\left | n+1 \right |\)

\(\displaystyle =\frac{1}{2}*\infty=\infty\).

Since \(\displaystyle \infty>1\),

we have sufficient evidence to conclude that the series is divergent.

In order to figure if 

\(\displaystyle \sum_{n=1}^{\infty} \frac{11^{n}}{(n+1)3^{n}}\)

is convergent, divergent or neither, we need to use the ratio test. 

Remember that the ratio test is as follows.

Suppose we have a series \(\displaystyle \sum a_n\). We define,

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

Then if 

\(\displaystyle L< 1\), the series is absolutely convergent.

\(\displaystyle L>1\), the series is divergent.

\(\displaystyle L=1\), the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply the ratio test to our problem.

Let  

\(\displaystyle a_n=\frac{11^n}{(n+1)3^{n}}\)

and

\(\displaystyle a_{n+1}=\frac{11^{n+1}}{(n+2)3^{n+1}}\).

Now 

\(\displaystyle L=\lim_{n\rightarrow \infty}\left | \frac{\frac{11^{n+1}}{(n+2)3^{n+1}}}{\frac{11^n}{(n+1)3^{n}}} \right |\)

\(\displaystyle =\lim_{n\rightarrow \infty}\left | \frac{(n+1)11^{n+1}3^n}{11^{n}(n+2)3^{n+1}} \right |\).

Now lets simplify this expression to 

\(\displaystyle =\lim_{n\rightarrow \infty}\left | \frac{11(n+1)}{3(n+2)} \right |\)

\(\displaystyle =\frac{11}{3}\lim_{n\rightarrow \infty}\left |\frac{(n+1)}{(n+2)} \right |\)

\(\displaystyle =\frac{11}{3}*1=\frac{11}{3}\).

Since \(\displaystyle \frac{11}{3}>1\),

we have sufficient evidence to conclude that the series is divergent.

To determine if

\(\displaystyle \sum_{n=1}^{\infty} \frac{n!\ 4^n}{n+1}\)

is convergent, divergent or neither, we need to use the ratio test.

The ratio test is as follows.

Suppose we a series  \(\displaystyle \sum a_n\). Then we define,

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

If

\(\displaystyle L< 1\)  the series is absolutely convergent (and therefore convergent).

\(\displaystyle L>1\)  the series is divergent.

\(\displaystyle L=1\) the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

\(\displaystyle a_n=\frac{n!\ 4^n}{n+1}\)

and

\(\displaystyle a_{n+1}=\frac{(n+1)!\ 4^{n+1}}{n+2}\)

Now

\(\displaystyle L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)!\ 4^{n+1}}{n+2}}{\frac{n!\ 4^n}{n+1}}\right |\)

We can rearrange the expression to be

\(\displaystyle L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)(n+1)!\ 4^{n+1}}{(n+2)n!\ 4^n}\right |\)

Now lets simplify this.

\(\displaystyle L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)(n+1) 4}{(n+2)}\right |\)

When we evaluate the limit, we get.

\(\displaystyle L=\infty\).

Since \(\displaystyle L>1\), we have sufficient evidence to conclude that the series diverges.

To determine if

\(\displaystyle \sum_{n=0}^{\infty} \frac{n \ 2^n}{5^n}\)

is convergent, divergent or neither, we need to use the ratio test.

The ratio test is as follows.

Suppose we a series  \(\displaystyle \sum a_n\). Then we define,

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

If

\(\displaystyle L< 1\)  the series is absolutely convergent (and thus convergent).

\(\displaystyle L>1\)  the series is divergent.

\(\displaystyle L=1\) the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

\(\displaystyle a_n=\frac{n \ 2^n}{5^n}\)

and

\(\displaystyle a_{n+1}=\frac{(n+1) \ 2^{n+1}}{5^{n+1}}\)

Now

\(\displaystyle L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1) \ 2^{n+1}}{5^{n+1}}}{\frac{n \ 2^n}{5^n}}\right |\)

We can rearrange the expression to be

\(\displaystyle L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)\ 2^{n+1}\ 5^n}{n\ 2^n \ 5^{n+1}}\right |\).

Now lets simplify this.

\(\displaystyle L=\lim_{n\rightarrow \infty}\left | \frac{2(n+1)}{5n}\right |\)

When we evaluate the limit, we get.

\(\displaystyle L=\frac{2}{5}\).

Since \(\displaystyle L< 1\), we have sufficient evidence to conclude that the series converges.

To determine if

\(\displaystyle \sum_{n=1}^{\infty} \frac{n(-100)^n}{(n+1)5^{n+2}}\)

is convergent, divergent or neither, we need to use the ratio test.

The ratio test is as follows.

Suppose we a series  \(\displaystyle \sum a_n\). Then we define,

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

If

\(\displaystyle L< 1\)  the series is absolutely convergent (therefore convergent).

\(\displaystyle L>1\)  the series is divergent.

\(\displaystyle L=1\) the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

\(\displaystyle a_n=\frac{n(-100)^n}{(n+1)5^{n+2}}\)

and

\(\displaystyle a_{n+1}=\frac{(n+1)(-100)^{n+1}}{(n+2)5^{n+3}}\)

Now

\(\displaystyle L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)(-100)^{n+1}}{(n+2)5^{n+3}}}{\frac{n(-100)^n}{(n+1)5^{n+2}}}\right |\)

We can rearrange the expression to be

\(\displaystyle L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2(-100)^{n+1}{5^{n+2}}}{n(n+2)\ (-100)^n \ 5^{n+3}}\right |\)

Now lets simplify this.

\(\displaystyle L=20\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2}{n(n+2)}\right |\)

When we evaluate the limit, we get.

\(\displaystyle L=20\).

Since \(\displaystyle L>1\), we have sufficient evidence to conclude that the series diverges.

To determine if

\(\displaystyle \sum_{n=0}^{\infty} 4^{2n+1}n!\)

is convergent, divergent or neither, we need to use the ratio test.

The ratio test is as follows.

Suppose we a series  \(\displaystyle \sum a_n\). Then we define,

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

If

\(\displaystyle L< 1\)  the series is absolutely convergent (and thus convergent).

\(\displaystyle L>1\)  the series is divergent.

\(\displaystyle L=1\) the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

\(\displaystyle a_n= 4^{2n+1}n!\)

and

\(\displaystyle a_{n+1}= 4^{2(n+1)+1}(n+1)!=4^{2n+3}(n+1)!\)

Now

\(\displaystyle L=\lim_{n\rightarrow \infty}\left | \frac{4^{2n+3}(n+1)!}{4^{2n+1}n!}\right |\)

We can simplify the expression to be

\(\displaystyle L=\lim_{n\rightarrow \infty}\left | 16 (n+1)}\right |\)

When we evaluate the limit, we get.

\(\displaystyle L=\infty\).

Since \(\displaystyle L>1\), we have sufficient evidence to conclude that the series diverges.

To determine whether this series is convergent, divergent or neither

\(\displaystyle \sum_{n=0}^{\infty} \frac{n4^{2n}}{(n+1)3^{2n}}\)

we need to remember the ratio test.

The ratio test is as follows.

Suppose we a series  \(\displaystyle \sum a_n\). Then we define,

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

If

\(\displaystyle L< 1\)  the series is absolutely convergent (and therefore convergent).

\(\displaystyle L>1\)  the series is divergent.

\(\displaystyle L=1\) the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

\(\displaystyle a_n= \frac{n4^{2n}}{(n+1)3^{2n}}\)

and

\(\displaystyle a_{n+1}= \frac{(n+1)4^{2(n+1)}}{(n+2)3^{2(n+1)}}= \frac{(n+1)4^{2n+2}}{(n+2)3^{2n+2}}\)

Now

\(\displaystyle L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)4^{2n+2}}{(n+2)3^{2n+2}}}{\frac{n4^{2n}}{(n+1)3^{2n}}}\right |\)

We can rearrange the expression to be

\(\displaystyle L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2\ 3^{2n} \ 4^{2n+2}}{n(n+2)\ 3^{2n+2}\ 4^{2n}}\right |\)

Now lets simplify this to.

\(\displaystyle L=\frac{16}{9}\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2}{n(n+2)}\right |\)

When we evaluate the limit, we get.

\(\displaystyle L=\frac{16}{9}\).

Since \(\displaystyle L>1\), we have sufficient evidence to conclude that the series is divergent.

To determine whether this series is convergent, divergent or neither

\(\displaystyle \sum_{n=0}^{\infty} \frac{n2^{4n}}{(n+1)5^{6n}}\)

we need to remember the ratio test.

The ratio test is as follows.

Suppose we a series  \(\displaystyle \sum a_n\). Then we define,

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

If

\(\displaystyle L< 1\)  the series is absolutely convergent (thus convergent).

\(\displaystyle L>1\)  the series is divergent.

\(\displaystyle L=1\) the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

\(\displaystyle a_n= \frac{n2^{4n}}{(n+1)5^{6n}}\)

and

\(\displaystyle a_{n+1}= \frac{(n+1)2^{4(n+1)}}{(n+2)5^{6(n+1)}}= \frac{(n+1)2^{4n+4}}{(n+2)5^{6n+6}}\)

Now

\(\displaystyle L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)2^{4n+4}}{(n+2)5^{6n+6}}}{\frac{n2^{4n}}{(n+1)5^{6n}}}\right |\)

We can rearrange the expression to be

\(\displaystyle L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2\ 2^{4n+4} \ 5^{6n}}{n(n+2)\ 2^{4n}\ 5^{6n+6}}\right |\)

Now lets simplify this to.

\(\displaystyle L=\frac{16}{15625}\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2}{n(n+2)}\right |\)

When we evaluate the limit, we get.

\(\displaystyle L=\frac{16}{15625}\).

Since \(\displaystyle L< 1\), we have sufficient evidence to conclude that the series is convergent.