Question

Determine the interval of convergence for the power series

$\displaystyle\sum_{n=1}^ ∞$ 3ⁿ(x-1)ⁿ/(n+1)²

Answer 1

Let $a_n=\dfrac{3^n(x-1)^n}{(n+1)^2}.$ Then

$$\bigg|\dfrac{a_{n+1}}{a_n}\bigg|=\bigg|\dfrac{\dfrac{3^{n+1}(x-1)^{n+1}}{(n+1+1)^2}}{\dfrac{3^n(x-1)^n}{(n+1)^2}}\bigg|$$

$$=3(\dfrac{n+1}{n+2})^2|x-1|\to3|x-1|\ as \ n\to\infin$$

By the Ratio Test, the given series converges if $3|x-1|<1$ and diverges if $3|x-1|>1.$

Thus it converges if $|x-1|<\dfrac{1}{3}$ and diverges if $|x-1|>\dfrac{1}{3}$ .

$$|x-1|<\dfrac{1}{3}$$

$$-\dfrac{1}{3}<x-1<\dfrac{1}{3}$$

$$\dfrac{2}{3}<x-1<\dfrac{4}{3}$$

This means that the series converges in the interval $(\dfrac{2}{3}, \dfrac{4}{3}).$

If $x=\dfrac{2}{3},$ the series becomes

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

The series $\displaystyle\sum_{n=1}^{\infin}\dfrac{1}{(n+1)^2}=\displaystyle\sum_{k=2}^{\infin}\dfrac{1}{k^2}$ converges as $p$ -series with $p=2>1.$

Then the series $\displaystyle\sum_{n=1}^{\infin}\dfrac{(-1)^n}{(n+1)^2}$ converges absolutely.

If $x=\dfrac{4}{3},$ the series becomes

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

The series $\displaystyle\sum_{n=1}^{\infin}\dfrac{1}{(n+1)^2}=\displaystyle\sum_{k=2}^{\infin}\dfrac{1}{k^2}$ converges as $p$ -series with $p=2>1.$

Therefore the interval of convergence is $\bigg[\dfrac{2}{3}, \dfrac{4}{3}\bigg].$