Question

Find the interval of convergence of the power series

∞

Σ [(−1)^𝑛 (𝑛 − 2) / 𝑛^2 . 2^𝑛] (𝑥 − 2)^𝑛

𝑛=3

Answer 1

$$\displaystyle\sum_{n=3}^{\infin}\dfrac{(-1)^n(n-2)}{n^22^n}(x-2)^n$$

$$\bigg|\dfrac{a_{n+1}}{a_n}\bigg|=\bigg|\dfrac{\dfrac{(n+1-2)}{(n+1)^22^{n+1}}(x-2)^{n+1}}{\dfrac{(n-2)}{n^22^n}(x-2)^n}\bigg|$$

$$=\dfrac{(n-1)n^2}{2(n-2)(n+1)^2}|x-2|\to\dfrac{1}{2}|x-2|\ as\ n\to\infin$$

$$\dfrac{1}{2}|x-2|<1=>|x-2|<2$$

Radius of convergence: $2.$

$x=0$

$$\displaystyle\sum_{n=3}^{\infin}\dfrac{(-1)^n(n-2)}{n^22^n}(0-2)^n=\displaystyle\sum_{n=3}^{\infin}\dfrac{n-2}{n^2}$$

Use the Limit Comparison Test with

$$\displaystyle\sum_{n=3}^{\infin}b_n=\displaystyle\sum_{n=3}^{\infin}\dfrac{1}{n}$$

$$\lim\limits_{n\to\infin}\dfrac{\dfrac{n-2}{n^2}}{\dfrac{1}{n}}=1>0$$

The garmonic series $\displaystyle\sum_{n=3}^{\infin}\dfrac{1}{n}$ diverges as $p$ series with $p=1.$

Then the series $\displaystyle\sum_{n=3}^{\infin}\dfrac{n-2}{n^2}$ diverges by the Limit Comparison Test.

$x=4$

$$\displaystyle\sum_{n=3}^{\infin}\dfrac{(-1)^n(n-2)}{n^22^n}(4-2)^n=\displaystyle\sum_{n=3}^{\infin}\dfrac{(-1)^n(n-2)}{n^2}$$

Use the Alternating Series Test

$$a_n=\dfrac{n-2}{n^2}$$

$$\lim\limits_{n\to\infin}a_n=\lim\limits_{n\to\infin}\dfrac{n-2}{n^2}=0$$

$$a_{n+1}=\dfrac{n+1-2}{(n+1)^2}<\dfrac{n-2}{n^2}=a_n, n\geq 3$$

Then the series $\displaystyle\sum_{n=3}^{\infin}\dfrac{(-1)^n(n-2)}{n^2}$ converges by the Alternating Series Test.

Therefore the interval of convergence is $(0, 4].$