Consider the power series $\sum_{n=2}^{\infty} (-1)^n(n+5)x^n$.
1. Show that the series converges for all $x\in (-1,1)$.
If we let $a_n = (-1)^n(n+5)x^n$ then using the ratio test we get
\begin{equation*}
\begin{split}
\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| &= \lim_{n\to\infty} \left|\frac{(-1)^{n+1}(n+6)x^{n+1}}{(-1)^n(n+5)x^n}\right| \\
&= |x|\lim_{n\to\infty} \left|-\frac{n+6}{n+5}\right| \\
&= |x|\lim_{n\to\infty} \frac{n+6}{n+5} \\
&= |x|.
\end{split}
\end{equation*}
So if $|x| < 1$, then this series will converge. Thus, the interval of convergence certainly contains $(-1,1)$. We need to check the convergence at the endpoints because the ratio test doesn't tell us anything if $\lim_{k\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = 1$. When $x = 1$, the series is $\sum_{n=1}^{\infty} (-1)^n(n+5)$. This obviously diverges as this is an alternating series. When $x = -1$, the series is $\sum_{n=1}^{\infty} n(n+5)$ which also diverges. Hence, the series converges for all $x\in (-1,1)$ as required.
Define the function $f:(-1,1)\mapsto \mathbb{R}$ by
\begin{equation*}
f(x) := \sum_{n=2}^{\infty} (-1)^n(n+5)x^n.
\end{equation*}
2. Give the fourth-order Taylor polynomial of $f$ about $0$.
This is given by
\begin{equation*}
f(x) = \sum_{n=2}^{5} (-1)^n(n+5)x^n = 7x^2-8x^3+9x^4-10x^5.
\end{equation*}
Define the function $g:(-1,1)\mapsto \mathbb{R}$ by
\begin{equation*}
g(x) := \int_{0}^{x} f(t) \; dt,
\end{equation*}
where $f$ is as above.
3. Find the Taylor series about $0$ for the function $g$.
We have
\begin{equation*}
\begin{split}
g(x) = \int_{0}^{x} f(t) \; dt &= \int_{0}^{x} \sum_{n=2}^{\infty} (-1)^n(n+5)t^n \; dt \\
&= \sum_{n=2}^{\infty} (-1)^n(n+5) \int_{0}^{x} t^n \; dt \\
&= \sum_{n=2}^{\infty} (-1)^n(n+5) \left[\frac{t^{n+1}}{n+1}\right]_{0}^{x} \\
&= \sum_{n=2}^{\infty} (-1)^n\frac{n+5}{n+1}x^{n+1}.
\end{split}
\end{equation*}
Does this look good. Any feedback would be appreciated!!!
In 2. your Taylor polynomial has degree $5$. You should give the fourth-order Taylor polynomial !
The rest looks fine !