What is the interval of convergence of the series sum_(n=0)^oo((-3)^n*x^n)/sqrt(n+1)?

Sep 27, 2014

The interval of convergence is $\left(- \frac{1}{3} , \frac{1}{3}\right]$.

Let ${a}_{n} = \frac{{\left(- 3\right)}^{n} {x}^{n}}{\sqrt{n + 1}}$. $R i g h t a r r o w {a}_{n + 1} = \frac{{\left(- 3\right)}^{n + 1} {x}^{n + 1}}{\sqrt{n + 2}}$.

By Ratio Test,

lim_{n to infty}|{a_{n+1}}/{a_n}| =lim_{n to infty}|{(-3)^{n+1}x^{n+1}}/{sqrt{n+2}}cdot{sqrt{n+1}}/{(-3)^nx^n}|

by cancelling out common factors,

$= {\lim}_{n \to \infty} | \frac{- 3 x \sqrt{n + 1}}{\sqrt{n + 2}} |$

by pulling $| - 3 x | = 3 | x |$ out of the limit,

$= 3 | x | {\lim}_{n \to \infty} \setminus \sqrt{\frac{n + 1}{n + 2}}$

by dividing the numerator and the denominator by $n$,

$= 3 | x | {\lim}_{n \to \infty} \setminus \sqrt{\frac{1 + \frac{1}{n}}{1 + \frac{2}{n}}} = 3 | x | \sqrt{\frac{1 + 0}{1 + 0}} = 3 | x | < 1$

By dividing by 3,

$R i g h t a r r o w | x | < \frac{1}{3} \Leftrightarrow - \frac{1}{3} < x < \frac{1}{3}$

So, the power series converges at least on $\left(- \frac{1}{3} , \frac{1}{3}\right)$.

Now, we need to check the endpoints.

When $x = - \frac{1}{3}$, the series becomes

${\sum}_{n = 0}^{\infty} \frac{{\left(- 3\right)}^{n} {\left(- \frac{1}{3}\right)}^{n}}{\sqrt{n + 1}} = {\sum}_{n = 0}^{\infty} \frac{1}{\sqrt{n + 1}} = {\sum}_{n = 1}^{\infty} \frac{1}{n} ^ \left\{\frac{1}{2}\right\}$,

which is a divergence p-series since $p = \frac{1}{2} \le 1$

So, $x = - \frac{1}{3}$ should be excluded.

When $x = \frac{1}{3}$, the serie becomes

${\sum}_{n = 0}^{\infty} \frac{{\left(- 3\right)}^{n} {\left(\frac{1}{3}\right)}^{n}}{\sqrt{n + 1}} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left\{\sqrt{n + 1}\right\}$,

which is a convergent alternating series by Alternating Series Test.

$\frac{1}{\sqrt{n + 1}} \ge \frac{1}{\sqrt{n + 2}}$ and ${\lim}_{n \to \infty} \frac{1}{\sqrt{n + 1}} = 0$

So, $x = \frac{1}{3}$ should be included.

Hence, the interval of convergence is $\left(- \frac{1}{3} , \frac{1}{3}\right]$.