You might already recognize this as the power series representation for \(\displaystyle e^x\). Since \(\displaystyle e^x\) is well defined for all \(\displaystyle x \in \mathbb{R}\), the radius of convergence is \(\displaystyle \infty\).

If we want to find the radius of covergence using convergence tests, we can use the Ratio Test here. We have-

\(\displaystyle \lim_{k\to\infty} \left |\frac{a_{k+1}}{a_k} \right | = \lim_{k\to\infty} \left | \frac{x^{k+1}}{(k+1)!} \frac{k!}{x^k} \right | =\lim_{k\to\infty} \left | \frac{x}{k+1}\right | =\)

\(\displaystyle \lim_{k\to\infty} \frac{\left | x \right |}{k+1} = \left | x \right | \lim_{k\to\infty} \frac{1}{k+1} = \left | x \right | \cdot 0 = 0\)

Since this limit equals \(\displaystyle 0\) regardless of the value of \(\displaystyle x\), and the Ratio Test indicates absolute convergence of a series when the above limit is less than \(\displaystyle 1\), \(\displaystyle \sum_{k=0}^{\infty} \frac{x^k}{k!}\) converges for all \(\displaystyle x \in \mathbb{R}\). Hence the radius of convergence is \(\displaystyle \infty\).

The question does not ask which function is the largest at any given point, it asks which grows fastest. For this question we need to look at all of the terms and determine which function has the dominant term. In this case \(\displaystyle x^x\) is the dominate term therefore, it will grow the fastest.

Write the correct definition of cosine as a power series.

\(\displaystyle cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...\)

Replace \(\displaystyle x\) with the term \(\displaystyle x^2\).

\(\displaystyle cos(x^2)=1-\frac{x^4}{2!}+\frac{x^8}{4!}-\frac{x^{12}}{6!}+...\)

The correct answer is:

\(\displaystyle 1-\frac{x^4}{2!}+\frac{x^8}{4!}-\frac{x^{12}}{6!}+...\)

With every power series, you must first start by using the ratio test.

The \(\displaystyle n+1\) series is \(\displaystyle \frac{x^{n+1}}{\ln n+1}\).

Using the ratio test you get 

\(\displaystyle \lim_{n \rightarrow \infty } \frac{x^{n+1}}{\ln n+1} \cdot \frac{\ln n}{x^{n}}\).

Canceling things out and taking the \(\displaystyle x\) out of the limit (as you are only taking the limit as n approaches infinity) gets you the following: 

\(\displaystyle x \cdot \lim_{n \rightarrow \infty } \frac{\ln n}{\ln n+1}\), which becomes x (as the limit is 1).

Because the ratio test states that the series can only converge if the absolute value of x is less than or equal to 1, this means that x is between -1 and 1. And so the radius of convergence is 1. 

Recall that a Taylor Series approximation at \(\displaystyle x=1\) is given by

\(\displaystyle f(x)= \sum_{n=0}^{\infty}\frac{f^{(n)}(1)}{n!}x^n\).

In our case, \(\displaystyle f(x)=e^{2x}\)

\(\displaystyle f^{(0)}(1)=e^2, f^{(1)}(1)=2e^2, f^{(0)}(1)=4e^2, f^{(3)}(1)=8e^2,...\)

So, our final summation will look like

\(\displaystyle e^{2x}= e^2\sum_{n=0}^{\infty}\frac{2^{n}}{n!}x^n=e^2\sum_{n=0}^{\infty}\frac{(2x)^{n}}{n!}\).

We can use the common Maclaurin Series expansion,

 \(\displaystyle e^u=\sum_{n=0}^{\infty}\frac{u^{n}}{n!}\)

and modify it to fit our current function \(\displaystyle xe^{x^2}\).

First, we substitute \(\displaystyle x^2\) in for \(\displaystyle u\). Doing so we obtain,

 \(\displaystyle e^{x^2}=\sum_{n=0}^{\infty}\frac{(x^2)^n}{n!}=\sum_{n=0}^{\infty}\frac{x^{2n}}{n!}\).

The remaining task is to get the \(\displaystyle x\) in front of the function. We do this by simply multiplying each side of the above equation by \(\displaystyle x\).

Doing so, we obtain,

 \(\displaystyle xe^{x^2}=x\sum_{n=0}^{\infty}\frac{x^{2n}}{n!}=\sum_{n=0}^{\infty}\frac{x^{2n}x}{n!}=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{n!}\).

Thus our Maclaurin Series for \(\displaystyle xe^{x^2}\) is,

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

Note: One could also try to use the Maclaurin Series formula \(\displaystyle f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n\) to compute the series directly. However, computing \(\displaystyle f^{(n)}(0)\) for this function for every \(\displaystyle n\) proves to be a formidable task. Therefore, it is easiest to solve these problems by remembering the Maclaurin Series for specific functions, and then using substitutions and multiplication tricks to build up to the function in question. I suggest memorizing the following Maclaurin Series':

\(\displaystyle e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}\)

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

\(\displaystyle \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}\)

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

There are a few others that are useful, but these are the most commonly used in classroom and exam settings.

\(\displaystyle \small \small f(x)=\frac{\sqrt{x}}{3-x}\)

To find a convergent power series representation of the function we can use the following theorem for an infinite geometric series that converges to a finite value: 

_________________________________________________________

\(\displaystyle \small \sum_{n=0}^{\infty}ar^n =\frac{a}{1-r}\)  for \(\displaystyle r\) such that \(\displaystyle \small \left | r\right |< 1\)

 ________________________________________________________

First factor the function to so that one of the factors is comparable to the form \(\displaystyle \frac{a}{1-r}\). 

\(\displaystyle \small \small \small \small \small f(x)=\sqrt{x}\left(\frac{1}{3-x} \right )=\sqrt{x}\left(\frac{1/3}{1-\frac{x}{3}} \right )\)

Now we can use the theorem to write the \(\displaystyle \frac{1/3}{1-\frac{x}{3}}\) factor as a convergent geometric series. Here our  "\(\displaystyle r\)" is just \(\displaystyle \frac{x}{3}\). Also, \(\displaystyle a = 1/3\). 

The series we will derive a power series that will converge to the factor \(\displaystyle \frac{1/3}{1-\frac{x}{3}}\). Also note that we do not have to "show" that \(\displaystyle \small \left | \frac{x}{3}\right | < 1\) to apply the theorem. We simply write it as an infinite series and then state that \(\displaystyle \small \left | \frac{x}{3}\right | < 1\) is a constraint on where the power series representation is valid. In other words, it's not a hypothesis we have to verify or check for. 

Using the theorem for a convergent infinite series, valid in this case for \(\displaystyle \small \left | \frac{x}{3}\right | < 1\), we can write the factor as follows: 

\(\displaystyle \small \small \small \small \left(\frac{1/3}{1-\left(\frac{x}{3}\right)} \right )=\sum_{n=0}^{\infty}\frac{1}{3}\left(\frac{x}{3} \right )^n\)

Now simply bring back the other factor, \(\displaystyle \sqrt{x}\), and the pull it into the summation. 

\(\displaystyle \small \small \small \small \small \sqrt{x}\small \small \left(\frac{1/3}{1-\left(\frac{x}{3}\right)} \right )=\sqrt{x}\sum_{n=0}^{\infty}\frac{1}{3}\left(\frac{x}{3} \right )^n = \sum_{n=0}^\infty \frac{1}{3^{n+1}}\sqrt{x} x^n = \sum_{n=0}^\infty \frac{1}{3^{n+1}} x^{\frac{1}{2}+n}\)

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

To find where the power series we've obtained converges, we simply have to apply the condition for the convergence of the geometric series \(\displaystyle \small \sum_{n=0}^{\infty}\frac{1}{3}\left(\frac{x}{3} \right )^n\). Even though the geometric series is not the actual power series we obtained for the function, it is still the series we based the power series representation on. All we did was multiply the function \(\displaystyle \sqrt{x}\) to each term in the summation. Therefore the interval of convergence- which is just the range over \(\displaystyle x\) for which the original function converges to the power series representation- must satisfy  \(\displaystyle \small \left | \frac{x}{3}\right |< 1\) 

This is equivalent to the interval  \(\displaystyle -3< x< 3\). Note that the interval is open at \(\displaystyle -3\) and \(\displaystyle 3\).

This is an indefinite integral so we can eliminate answers without the +C.

First, we need to find a power series representation for sin(x), which is 

\(\displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{(2k+1)!}\).

We then substitute in \(\displaystyle x^{3}\) to get

 \(\displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k (x^{3})^{2k+1}}{(2k+1)!}\) and then integrate the general term \(\displaystyle (x^3)^{2k+1}=x^{6k+3}\).

\(\displaystyle \int x^{6k+3}\) becomes \(\displaystyle \frac{x^{6k+3+1}}{6k+3+1}=\frac{x^{6k+4}}{6k+4}\) to get the correct answer, remembering our +C.

\(\displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k (x^{6k+4})}{(6k+4)(2k+1)!}+C\)