These are questions given for an assignment and I REALLY need to get a good grade so please could you guys check over them and give some pointers where I've gone wrong?
thanks.
ed: I hope it's ok to overwrite the original post:
$$\sum_0^\infty {\frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}}$$
\begin{align} a_n &= \frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}\\ a_{n+1} &= \frac{(2x-8)^{n+1}}{4^{n+1}((n+1)^{2}+2)}.\\ \left|\frac{a_n}{a_{n+1}}\right| &= \left|\frac{(2x-8)^{n+1}}{4^{n+1}((n+1)^{2}+2)} \times \frac{4^{n}(n^{2}+2)}{(2x-8)^{n}}\right|.\\ \lim_{n \to \infty}\left|\frac{a_n}{a_{n+1}}\right|&= |2x-8| \lim_{n \to \infty}\frac{n^{2}+2}{4(n^{2}+2n+3)}\\ &= |2x-8| \times\frac{1}{4}. \end{align} Therefore $L = \frac{1}{4}$ and series converges when $|2x-8| \times\frac{1}{4} < 1$.
Radius of convergence is $2$.
Endpoints; $|2x-8| = 4 \rightarrow x = 6$ and $x=2$.
Testing endpoints. When $x = 6$: $$\sum_0^\infty {\frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}} = \sum_0^\infty \frac{1}{n^{2}+2}.$$ convergent by comparison test with $\sum\frac{1}{n^{2}}$.
When x = 2: $$\sum_0^\infty {\frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}} = \sum_0^\infty\frac{(-1)^n}{n^{2}+2},$$ convergent by alternating sequence test and is absolutely convergent.
Therefore the series $$\sum_0^\infty {\frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}}$$ is convergent for $x\in [2,6]$.
2)b
$$\sum_1^\infty\frac{(-1)^n}{n4^{2n}}(x+3)^n$$
\begin{align} a_n &= \frac{(-1)^n}{n4^{2n}},\\ a_{n+1} &= \frac{(-1)^{n+1}}{(n+1)4^{2(n+1)}}.\\ \left|\frac{a_{n+1}}{a_n}\right| &= \left|\frac{(-1)^{n+1}}{(n+1)4^{2(n+1)}} \times \frac{n4^{2n}}{(-1)^n}\right|\\ \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| &= |x+3|\lim_{n \to \infty}\frac{n}{4^2(n+1)}\\ &= |x+3|\times\frac{1}{16}. \end{align} Series converges when $|x+3|\times\frac{1}{16} \lt 1$. Radius of convergence \is $16$.
Endpoints: $|x+3| = 16 \rightarrow x = 13$ or $-19$.
Testing endpoints: When $x = 13$, $$\sum_1^\infty\frac{(-1)^n(16)^n}{n4^{2n}} = \sum_1^\infty\frac{(-1)^n)}{n}.$$ This is an alternating harmonic series and therefore convergent.
When x = -19, $$\sum_1^\infty\frac{(-1)^n(-16)^n}{n4^{2n}}{} = \sum_1^\infty\frac{(-1)^{2n}}{n}.$$ Harmonic series and therefore divergent.
Therefore the series $$\sum_1^\infty\frac{(-1)^n}{n4^{2n}}(x+3)^n$$ is convergent for $x \in(-19, 13]$.
2) c)
$$\sum_0^\infty\frac{(1+3^{n-1})x^n}{(n+1)!}$$ \begin{align} a_n &= \frac{(1+3^{n-1})x^n}{(n+1)!}\\ a_n+1 &= \frac{(1+3^{(n+1)-1})x^{n+1}}{((n+1)+1)!}.\\ \left|\frac{a_{n+1}}{a_n}\right| &= \left|\frac{(1+3^n)x^{n+1}}{(n+2)!} \times \frac{(n+1)!}{(1+3^{n-1})x^n}\right|\\ \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| &= |x|\lim_{n \to \infty} \frac{1+3^n}{1+3^{n-1}} \times \lim_{n \to \infty} \frac{1}{n+2}\\ &= |x|\lim_{n \to \infty} \frac{3^{-n}+ 1}{3^{-n}+3^{-1}} \times 0\\ &= |x|\lim_{n \to \infty} \frac{0 + 1}{0 +3^{-1}} \times 0\\ &= |x|\times3\times0 = 0. \end{align}
Therefore, $L = 0$ and the series converges absolutely for all $x$. The radius of convergence is $\infty$.
$\displaystyle\sum_0^\infty\frac{(1+3^{n-1})x^n}{(n+1)!}$ converges absolutely for $x\in(-\infty,\infty)$.
Cheers,
Gregg