To determine the convergence or divergence of this series, we use the Ratio Test:

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

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

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

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

So, we evaluate the limit according to the formula above:

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

which simplified becomes

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

Further simplification results in

\(\displaystyle \lim_{n\rightarrow \infty }\left |\frac{2}{n+1} \right |=0< 1\)

Therefore, the series is absolutely convergent.

To determine if

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

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 hence 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

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

and

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

Now

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

We can rearrange the expression to be

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

We can simplify the expression to

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

When we evaluate the limit, we get.

\(\displaystyle L=2\).

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

To determine if

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

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 hence 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

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

and

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

Now

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

We can rearrange the expression to be

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

We can simplify the expression to

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

When we evaluate the limit, we get.

\(\displaystyle L=\frac{2}{3}\).

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

To determine if

\(\displaystyle \sum_{n=0}^{\infty} \frac{(n+1)!(-5)^n}{n!\2^{n+1}}\)

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 hence 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

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

and

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

Now

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

We can rearrange the expression to be

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

We can simplify the expression to

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

When we evaluate the limit, we get.

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

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

To determine if

\(\displaystyle \sum_{n=0}^{\infty} \frac{(2n+3)4^{n+2}}{(n+2)5^{n+1}}\)

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 hence 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

\(\displaystyle a_n=\frac{(2n+3)4^{n+2}}{(n+2)5^{n+1}}\)

and

\(\displaystyle a_{n+1}=\frac{(2(n+1)+3)4^{n+3}}{(n+3)5^{n+2}}=\frac{(2n+5)4^{n+3}}{(n+3)5^{n+2}}\)

Now

\(\displaystyle L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(2n+5)4^{n+3}}{(n+3)5^{n+2}}}{\frac{(2n+3)4^{n+2}}{(n+2)5^{n+1}}}\right |\)

We can rearrange the expression to be

\(\displaystyle L=\lim_{n\rightarrow \infty}\left | \frac{(2n+5)(n+2)4^{n+3}\ 5^{n+1}}{(n+3)(2n+3)5^{n+2}\ 4^{n+2}}\right |\)

We can simplify the expression to

\(\displaystyle L=\frac{4}{5}\lim_{n\rightarrow \infty}\left | \frac{(2n+5)(n+2)}{(n+3)(2n+3)}\right |\)

When we evaluate the limit, we get.

\(\displaystyle L=\frac{4}{5}\).

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

We know that:

\(\displaystyle chn=\frac{e^n+e^{-n}}{2}\) and therefore we deduce :

\(\displaystyle ch2n=\frac{e^{2n }+e^{-2n}}{2}\)

We will use the Comparison Test with this problem. To do this we will look at the function in general form \(\displaystyle v_{n}\equiv \frac{e^n} {e^{2n}}\). 

We can do this since,

\(\displaystyle e^{-n}=\frac{1}{e^n}\) and  \(\displaystyle e^{-2n}=\frac{1}{e^{2n}}\) approach zero as n approaches infinity. The limit of our function becomes,

\(\displaystyle \lim_{n\rightarrow \infty}\frac{e^n+e^{-n}}{e^{2n}+e^{-2n}}=\frac{e^n}{e^{2n}}\)

This last part gives us \(\displaystyle v_{n}\equiv {e^{-n}}\).

Now we know that \(\displaystyle \frac{1}{e}< 1\) and noting that \(\displaystyle e^{-n}}\) is a geometric series that is convergent.

We deduce by the Comparison Test that the series

having general term \(\displaystyle v_{n}\) is convergent.

We will use the comparison test to prove this result. We must note the following:

\(\displaystyle \frac{n^3+1}{n^4}\)  is positive.

We have all natural numbers n:

\(\displaystyle n^4\ge n^3\) , this implies that

\(\displaystyle n^4\ge n^3 \ge n\).

Inverting we get :

\(\displaystyle \frac{1}{n} < \frac{n^3+1}{n^4}\)

Summing from 1 to \(\displaystyle \infty\), we have

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n} < \sum_{n=1}^{\infty}\frac{n^3+1}{n^4}\)

We know that the \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}\) is divergent. Therefore by the comparison test:

\(\displaystyle \sum_{n=1}^{\infty}\frac{n^3}{n^4+1}\)

is divergent

We will use the Limit Comparison Test to study the nature of the series.

We note first that \(\displaystyle n\ge 2\), the series is positive.

We will compare the general term to \(\displaystyle \frac{1}{n^8}\). 

We note that by letting \(\displaystyle v_{n}=\frac{n+11}{n^9-7}\) and \(\displaystyle u_{n}=\frac{1}{n^8}\), we have:

\(\displaystyle \lim_{n\rightarrow \infty}\frac{u_{n}}{v_n}}=1\).

Therefore the two series have the same nature, (they either converge or diverge at the same time). 

We will use the Integral Test to deduce that the series having the general term:

\(\displaystyle u_{n}=\frac{1}{n^8}\) is convergent.

Note that we know that \(\displaystyle \int_{1}^{\infty}\frac{1}{x^p}dx\)is convergent if p>1 and in our case p=8 .

This shows that the series having general term \(\displaystyle u_{n}\) is convergent.

By the Limit Test, the series having general term \(\displaystyle v_n\) is convergent.

This shows that our series is convergent.

We will use the Comparison Test to prove this result. We must note the following:

\(\displaystyle \frac{n}{n^2+1}\)  is positive. 

We have all natural numbers n:

\(\displaystyle n^2\ge n\) , this implies that

\(\displaystyle n^2+1\ge n+1 > n\).

Inverting we get :

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

Summing from 1 to \(\displaystyle \infty\), we have

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

We know that the \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}\) is divergent. Therefore by the Comparison Test:

\(\displaystyle \sum_{n=1}^{\infty}\frac{n}{n^2+1}\) is divergent.

We will use the comparison test to prove that

\(\displaystyle \small \sum_{n=1}^\infty \frac{n^3}{n^7+1000}\)

converges (Note: we cannot use the ratio test, because then the ratio will be \(\displaystyle \small 1\), which means the test is inconclusive).

We will compare \(\displaystyle \small \frac{n^3}{n^7+1000}\) to \(\displaystyle \small \frac{1}{n^4}\) because they "behave" somewhat similarly. Both series are nonzero for all \(\displaystyle \small n\geq 1\), so one of the conditions is satisfied.

The series 

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

converges, so we must show that 

\(\displaystyle \small \small \frac{1}{n^4}\geq \frac{n^3}{n^4+1000}\) 

for \(\displaystyle \small n\geq 1\).

This is easy to show because

\(\displaystyle \small \frac{1}{n^4}=\frac{n^3}{n^7}\geq \frac{n^3}{n^7+1000}\)

since the denominator \(\displaystyle \small n^7+1000\) is greater than or equal to \(\displaystyle \small n^7\) for all \(\displaystyle \small n\geq 1\).

Thus, since 

\(\displaystyle \small \small \frac{1}{n^4}\geq \frac{n^3}{n^4+1000}\)

and because

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

converges, it follows that 

\(\displaystyle \small \sum_{n=1}^\infty \frac{n^3}{n^7+1000}\)

converges, by comparison test.