Solutions - 0220-02

(a)

$$\begin{array}{c}|\frac{{(x+3)}^{n+1}}{(n+1)!}\cdot \frac{n!}{{(x+3)}^n}|\gg \gg \frac{1}{n+1}|x+3|\stackrel{n\to \infty }{⟶}0\end{array}$$

Therefore $R=\infty$ and $I=(-\infty ,+\infty )$.

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(b)

$$|\frac{{(x-7)}^{n+1}}{n+1}\cdot \frac{n}{{(x-7)}^n}|\gg \gg \frac{n}{n+1}|x-7|\stackrel{n\to \infty }{⟶}0$$

Therefore $R=1$ and the preliminary interval is $(\mathrm{6,8})$.

At $x=6$ we have ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{{(-1)}^n}{n}=\sum _{n=0}^{\infty }\frac{1}{n}}$. This diverges ($p=1$).

At $x=8$ we have ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{{(+1)}^n}{n}=\sum _{n=0}^{\infty }\frac{{(-1)}^n}{n}}$. This converges by the AST.

Therefore, the final interval of convergence is $I=(\mathrm{6,8}]$.

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(c)

$$\left|\frac{{(n+1)}^{n+1}{(x-2)}^{n+1}}{n^n{(x-2)}^n}\right|\gg \gg (n+1){(1+\frac{1}{n})}^n|x-2|$$

Observe that $1+\frac{1}{n}>1$ so ${\lim }_{n\to \infty }{(1+\frac{1}{n})}^n\ge 1$. Assume $x\ne 2$. Then:

$$(n+1){(1+\frac{1}{n})}^n|x-2|\stackrel{n\to \infty }{⟶}\infty$$

If $x=2$, then of course the series is $\sum 0$ and converges to $0$.

Therefore $R=0$ and $I=\{2\}$.