Radius and Interval of Convergence

**schnek** · November 24th 2009, 02:52 PM

Determine the radius and interval of convergence of the series ∑(n=0, ∞) [n(x+2)^n]/[3^(n+1)]

**Soroban** · November 24th 2009, 03:30 PM

Hello, schnek!

Determine the radius and interval of convergence of the series"

. . . $\sum^{\infty}_{n=0} \frac{n(x+2)^n}{3^{n+1}}$

Use the Ratio Test . . .

$R \;=\;\frac{a_{n+1}}{a_n} \;=\; \frac{(n+1)(x+2)^{n+1}}{3^{n+2}}\cdot \frac{3^{n+1}}{n(x+2)^n} \;=\;\frac{n+1}{n}\cdot\frac{(x+2)^{n+1}}{(x+2)^n} \cdot\frac{3^{n+1}}{3^{n+2}}$

. . . $=\frac{n+1}{n}\cdot(x+2)\cdot\frac{1}{3} \;=\;\frac{x+2}{3}\cdot\frac{n+1}{n}\;=\;\frac{x+2 }{3}\left(1 + \frac{1}{n}\right)$

Then: . $\lim_{n\to\infty} |R| \;=\;\lim_{n\to\infty} \left|\frac{x+2}{3}\cdot\left(1 + \frac{1}{n}\right)\right| \;=\;\left|\frac{x+2}{3}\right|$

So we have: . $\left|\frac{x+2}{3}\right| \:<\:1 \quad\Rightarrow\quad |x + 2| \:<\:3 \quad\hdots\quad \boxed{\text{ radius of convergence: 3 }}$

. . $-3 \:<\:x + 2 \:<\:3 \quad\Rightarrow\quad -5 \:<\:x \:<\:1$

At $x = \text{-}5$ , we have: . $\sum^{\infty}_{n=0}\frac{n(\text{-}3)^n}{3^{n+1}} \;=\;\sum^{\infty}_{n=0}\frac{(\text{-}1)^n n}{3}\;=\;\tfrac{1}{3}\sum^{\infty}_{n=0}(\text{-}1)^n n$ . . . diverge

At $x = 1$ , we have: . $\sum^{\infty}_{n=0}\frac{n\!\cdot\!3^n}{3^{n+1}} \;=\; \sum^{\infty}_{n=0}\frac{n}{3} \;=\;\tfrac{1}{3}\sum^{\infty}_{n=0}n$ . . . diverge

. . $\boxed{\text{ Interval of convergence: }\;(-5,\:1)\;}$

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