Solutions - 0220-03

(a)

Apply the ratio test:

$$\begin{array}{c}|a_{n+1}\cdot a_n^{-1}|=|\frac{{(x-8)}^{n+1}}{{(n+1)}^4+1}\cdot \frac{n^4+1}{{(x-8)}^n}|\\ \\ \gg \gg \frac{n^4+1}{{(n+1)}^4+1}|x-8|\\ \\ \gg \gg \frac{n^4+1}{n^4+4n^3+6n^2+4n+1}|x-8|\stackrel{n\to \infty }{⟶}|x-8|\end{array}$$

Therefore $R=1$ and the preliminary interval is $(\mathrm{7,9})$.

Check endpoints:

At $x=7$, we have ${\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n}{n^4+1}}$, which converges absolutely.

At $x=9$, we have ${\displaystyle \sum _{n=0}^{\infty }\frac{1}{n^4+1}}$, which converges by the DCT, comparing with ${\displaystyle \sum \frac{1}{n^4}}$.

Therefore, the final interval of convergence is $I=[\mathrm{7,9}]$.

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(b)

Apply the ratio test:

$$\begin{array}{c}|a_{n+1}\cdot a_n^{-1}|=|\frac{x^{n+1}}{3\cdot 7\cdot 11\cdots (4n-1)(4n+3)}\cdot \frac{3\cdot 7\cdot 11\cdots (4n-1)}{x^n}|\\ \\ \gg \gg \frac{1}{4n+3}|x|\stackrel{n\to \infty }{⟶}0\end{array}$$

Therefore, $R=\infty$ and the interval of convergence is $I=(-\infty ,+\infty )$.