a) To prove that the sequence $a_n=\frac{c^n}{n!}$ is decreasing for $n\gg1$, we need to show that $a_{n+1} < a_n$ for all $n\gg1$:

\begin{align*} a_{n+1} & = \frac{c^{n+1}}{(n+1)!} \ & = \frac{c}{n+1} \cdot \frac{c^n}{n!} \ & < \frac{c}{n+1} \cdot \frac{c^n}{n!} \quad \text{since } c>1 \ & = \frac{c}{n} \cdot \frac{c^n}{n!} \cdot \frac{n}{n+1} \ & < \frac{c}{n} \cdot \frac{c^n}{n!} \quad \text{since } \frac{n}{n+1} < 1 \ & = a_n \end{align*}

Therefore, $a_n$ is a decreasing sequence for $n\gg1$.

b) To prove that $\lim_{n\rightarrow\infty}a_n=0$, we will use Theorem 3.4 which states that if there exist constants $C>0$ and $p>0$ such that $|a_{n+1}/a_n|\leq Cn^{-p}$ for all $n\geq N$, where $N$ is some suitable point in the sequence, then $\sum_{n=1}^{\infty}a_n$ converges absolutely.

Let us first find a suitable point $N$ in the sequence. Since $n!$ grows faster than any polynomial in $n$, we expect that $a_n$ will converge to zero faster than $1/n!$. To confirm this, we can take the ratio of consecutive terms:

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

So we have $|a_{n+1}/a_n|\leq 1/2$, which means that we can take $C=1/2$. Also, we can take $p=1$ since $1/n$ is a decreasing function for $n\geq 1$. Thus, we have $|a_{n+1}/a_n|\leq Cn^{-p}$ for all $n\geq 1$ with $C=1/2$ and $p=1$.

Now we need to find a suitable point $N$ in the sequence. Since $n!$ grows faster than any polynomial in $n$, we expect that $a_n$ will converge to zero faster than $1/n!$. To confirm this, we can solve the inequality $1/2^n\leq \epsilon$, where $\epsilon>0$ is a small number that we want to be close to zero. Taking the logarithm of both sides, we get $n\ln(2)\leq \ln(1/\epsilon)$, which implies that $n\geq (\ln(1/\epsilon))/\ln(2)$. Therefore, we can take $N=\lceil(\ln(1/\epsilon))/\ln(2)\rceil$, where $\lceil x\rceil$ denotes the smallest integer greater than or equal to $x$.

With these values of $C$, $p$, and $N$, we can use Theorem 3.4 to conclude that $\sum_{n=1}^{\infty}a_n$ converges absolutely, which implies that $\lim_{n\rightarrow\infty}a_n=0$.

c) Another way to prove that $\lim_{n\rightarrow\infty}a_n=0$ is to consider the series $\sum_{n=0}^{\infty}a_n$. We can use the ratio test to show that this series converges:

$\lim {n \rightarrow \infty} \frac{a{n+1}}{a_n}=\lim {n \rightarrow \infty} \frac{(n+1) !}{2^{n+1} n !}=\lim {n \rightarrow \infty} \frac{1}{2}=\frac{1}{2}<1$

Therefore, the series $\sum_{n=0}^{\infty}a_n$ converges, which implies that $\lim_{n\rightarrow\infty}a_n=0$ by the nth-term test for convergence.

We can apply the ratio test to determine the radius of convergence of the given series. Let

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

Then, the ratio of successive terms is

\begin{align*} \frac{a_{n+1}}{a_n} &= \frac{(2(n+1))!x^{n+1}}{(n+1)!(n+2)!} \cdot \frac{n!(n+1)!}{(2n)!x^n} \ &= \frac{(2n+2)(2n+1)x}{(n+2)(n+1)} \ &= \frac{4x^2}{(n+2)(n+1)} \cdot \frac{(2n+1)(2n+2)}{4(n+1)} \ &= \frac{4x^2}{(n+2)(n+1)} \cdot \left(1 + \frac{1}{n+1}\right) \cdot \left(1 + \frac{2}{n+1}\right). \end{align*}

Taking the limit as $n \to \infty$, we get

$\lim {n \rightarrow \infty} \frac{a{n+1}}{a_n}=\lim _{n \rightarrow \infty} \frac{4 x^2}{(n+2)(n+1)} \cdot\left(1+\frac{1}{n+1}\right) \cdot\left(1+\frac{2}{n+1}\right)=4 x^2$.

Therefore, the series converges absolutely if $4x^2 < 1$, or equivalently, $|x| < \frac{1}{2}$. On the other hand, the series diverges if $4x^2 > 1$, or equivalently, $|x| > \frac{1}{2}$. Thus, the radius of convergence is $R = \frac{1}{2}$.

To show that $1/k$ is a cluster point of the sequence ${h(n)/s(n)}$ for every positive integer $k$, we need to show that for any $\epsilon>0$ and positive integer $k$, there exist infinitely many $n$ such that $|h(n)/s(n)-1/k|<\epsilon$.

Let $\epsilon>0$ and $k$ be given. Choose $p$ to be any prime greater than $k/\epsilon$, and consider the numbers $n=p^2,p^3,\ldots,p^{k+1}$. Note that $h(n)=p$ and $s(n)=1+p+p^2+\cdots+p^k=\frac{p^{k+1}-1}{p-1}$.

For any $n=p^m$ with $2\leq m\leq k+1$, we have

$\left|\frac{h(n)}{s(n)}-\frac{1}{k}\right|=\left|\frac{k p-p^{k+1}}{(p-1)\left(p^{k+1}-1\right)}\right|=\frac{p\left(k-p^k\right)}{(p-1)\left(p^{k+1}-1\right)}<\frac{p(k-p)}{(p-1) p^{k+1}}<\frac{k}{p^k-1}<\epsilon$,

where we used the fact that $p>k/\epsilon$ and $p^{k+1}>kp$. Thus, we have shown that for any $\epsilon>0$ and positive integer $k$, there exist infinitely many $n$ such that $|h(n)/s(n)-1/k|<\epsilon$. Hence, $1/k$ is a cluster point of the sequence ${h(n)/s(n)}$ for every positive integer $k$.

To show that the sequence ${h(n)/s(n)}$ has no limit, we need to find two increasing sequences of integers ${a_n}$ and ${b_n}$ such that $h(a_n)/s(a_n)$ and $h(b_n)/s(b_n)$ have different limits as $n\rightarrow\infty$.

Let ${p_n}$ be the sequence of prime numbers, and define $a_n=p_{2n-1}^2$ and $b_n=p_{2n}^2$. Then we have

$\frac{h\left(a_n\right)}{s\left(a_n\right)}=\frac{p_{2 n-1}}{1+p_{2 n-1}} \rightarrow 1, \quad \frac{h\left(b_n\right)}{s\left(b_n\right)}=\frac{p_{2 n}}{1+p_{2 n}+p_{2 n}^2} \rightarrow 0$,

as $n\rightarrow\infty$, since $p_{2n-1}/(1+p_{2n-1})\rightarrow 1$ and $p_{2n}/(1+p_{2n}+p_{2n}^2)\rightarrow 0$. Therefore, the sequence ${h(n)/s(n)}$ does not have a limit as $n\rightarrow\infty$.

Since $f(x)$ has three distinct zeros on the interval $I$, we know that $f(x)$ changes sign at each of these zeros. Without loss of generality, assume that $f(x)$ is positive on $(a_0, a_1)$, negative on $(a_1, a_2)$, and positive on $(a_2, b)$.

Since $f(x)$ has a local extremum at $x=a_2$ (because $f^{\prime}(a_2)=0$), the second derivative $f^{\prime \prime}(x)$ must change sign at $x=a_2$. Specifically, $f^{\prime \prime}(x)$ is negative on $(a_1, a_2)$ and positive on $(a_2, b)$.

Now consider the third derivative $f^{\prime \prime \prime}(x)$. Since $f^{\prime \prime}(x)$ changes sign at $x=a_2$, we know that $f^{\prime \prime \prime}(x)$ must have a local extremum at $x=a_2$. Without loss of generality, assume that $f^{\prime \prime \prime}(x)$ is negative on $(a_1, a_2)$ and positive on $(a_2, b)$.

Since $f(x)$ changes sign at each of the zeros $a_0<a_1<a_2$, we know that $f(x)$ must have at least one local extremum (a maximum or minimum) between each pair of zeros. Specifically, $f(x)$ has a local minimum at some point $c_1$ in $(a_0, a_1)$, a local maximum at some point $c_2$ in $(a_1, a_2)$, and a local minimum at some point $c_3$ in $(a_2, b)$.

Since $f^{\prime \prime \prime}(x)$ is negative on $(a_1, a_2)$ and positive on $(a_2, b)$, we know that $f^{\prime \prime \prime}(c_2)>0$ and $f^{\prime \prime \prime}(c_3)<0$. Therefore, by the intermediate value theorem, there must be some point $c$ in $(c_2, c_3)$ such that $f^{\prime \prime \prime}(c)=0$, as desired.